Need advice ... What settings for a pellet boiler?

[Did67]

Yes, the room sensor shifts on a parallel. But suddenly, your foot is wrong.

[Phil532140]

Dirk,
I'm not very good at maths ...
Can you give me the calculation formula to find the theoretical water T ° from these values:
[A] = -10 (mini winter t °)
[B] = 79 (t ° of water at -10)
[C] = 19 (min summer t)
[D] = 36 (t ° of the water for 19 °)
[F] = 20 (no heating temperature)
[G] = 25 (min t ° of water)

Thank you for your help.
Philippe.

[dirk pitt]

second math course: equation of a straight line.
coef director (slope) = m = (Yb-Ya) / (Xb-Xa)
intercepted = Ya-mXa
in your case: slope = -1.48
origin = 64.1
with that, you can calculate the theoretical start time for all ext temp.
for example: if temp ext = 5 °
start time = -1.48 * 5 + 64.1 = 56.7

[Phil532140]

Thank you it's true that in the second I preferred English
Aie aie ... The maths and me ... on the ordinate at the origin I find 50.83 and not 64.1, there's something I did not get it?
And the set temperature does not enter the formula?
Phil.

[Phil532140]

And while I am there, I will also like to understand the role of the minimum water temperature in the calculation of the flow temperature. If we increase this min T ° we increase the starting t °? This is the parallel?
Goods.
Philippe.

[Did67]

To my knowledge, she does not intervene.

Min temp, like "max" temp, are in fact limits. If the calculation gives values ​​going above or below, the value is not taken into account. When the limits are reached, either the start is limited to the maximum (eg to prevent heated floors from "overheating"). Either the heating stops (mini), because we estimate that we are wasting energy (circulators, etc.) for something that is no longer useful

[Phil532140]

Thank you Did67 for this answer. Dirk pitt can you tell me how you find 64.1?
And why the setpoint does not enter the calculation?
Unless I do not understand everything ;-)
Thank you very much.
Philippe.

[sicetaitsimple]

You can write it differently:

Tdpart = 36 + 1,48 * (19-Text)

The set temperature (19 or 20 ° C that you wish to have inside), or rather the difference between the temperature observed and this set temperature, intervenes in correction of this basic curve, in more or less . That's why it's best to disable it to do the exercise suggested above by Dirk Pitt.

[Phil532140]

Ok in this case for 2.9 ° outside, I must have for a slope of 1.43 with 18 ° requested:

36+1.43*(18-2.9) = 58.34

Off the boiler sends me a starting T ° calculated 44 ° ... But the house is good to 18 ° ... I do not understand

[Phil532140]

Thank you all.
In fact, I really have the impression that my boiler is not powerful enough. Still unable to deliver 60 ° (Departure setpoint) this morning (it tops out at 47 ° at the start) ... In addition, after a few cycles it stops (It displays STOP ... It must estimate that the difference between the Departure temperature and return temperature allow it and that she can "live" on her reserve?) Do you have any idea?
Otherwise ... In short we will do with waiting to change or find a complementary solution when the T ° is less than 5 °