2
$\begingroup$

I have a thermometer in my room that takes readings over time and stores them into database. I also have a thermometer on the outside that is keeping track of the outdoor air temperature.

Let's say my reading are as follow: There are not heat radiating bodies in the room at the current moment. In two hours of passed time the room temperature have fallen from 20.3 to 20 degrees Celsius. The outdoor temperature was the same during that time and was 0 degrees Celsius.

Is it possible to calculate "good enough" heat loss number of a room only by this readings? I would be interested in using this calculations to predict the needed power requirements in order to keep particular room at 20.3 degrees for the next 8 hours.

EDIT: The volume of the room is 30m³.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

What do you mean by heat loss number? Depending on your definition, it might be necessary to know the cubic volume of the room.

1) Specific Heat Capacity of air is approx. 1.006 kJ/kgC

2) Air density (assuming dry air) is approx 1.2574 kg/m3

3) Mass = density x volume

4) Temperature differential is 0.3C

Gives us the following equation: heat capacity x mass x temperature differential = heat in KJ

For the second part of your question, you simply need to figure out the heat loss over 8 hours and supply that much heat to the air.

Som practical issues that will complicate things:

1) Outside air temperature. The greater the difference, the faster the heat transfer (into or out of the room).

2) Solar heating of room's exterior walls.

3) Calculating the amount of heat your system actually adds to the room air temperature.

4) Mixing the air to ensure it is evenly heated.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Well this should be some coefficient or a number that I could use to solve the second part of my "problem". I have added the volume figure. $\endgroup$
    – Denis Vitez
    Commented Feb 5, 2017 at 22:40
  • $\begingroup$ I want to thank you for your answer Jake. So how could I use the outside temperature to better predict my temperature loss? I was thinking of doing multiple readings (with different outside and inside temperatures) and then trying to fit a curve onto them. But this process will take me a long time so maybe there is a simpler way? $\endgroup$
    – Denis Vitez
    Commented Feb 6, 2017 at 15:06
  • $\begingroup$ It's going to be a negative curve for cooling, but it will be proportionate to the difference in temperature. I'd look up Fourier's Law or Newton's Cooling Law for more detailed explanations, but start with this to calculate heat loss: Q=UAΔT where A is the size of your area, U is the inverse of the room's R (insulation) value, and delta T is the temperature differential. Windows, degradation of insulation, convection, temperature in adjoining rooms, and many other things will throw off your real-world application some, but the calculation should be close. $\endgroup$
    – Jake Watrous
    Commented Feb 6, 2017 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.