How room temperature will change in 1 hour

[baracoda]

Hello, having the following data can I somehow calculate how will the temperature in room change in 1 hour time?

room length = 6m
room width = 6m
room height = 3m

lets say radiator is 0.5m x 1.5m and has 2000W and it can be in 2 states: on/off

I have also the current air temperature and relative humidity.

It is just a hipothetical question so I can assume some needed parameters if necessary.

 

[mfb]

It will depend on the walls and the interior.
Air has a specific heat capacity of 1J/(g*K), heating it alone with 2000 W would create a sauna. That's not what happens, of course. The solid objects in the room will absorb most of the heat.

 

[russ_watters]

If you use the specific heat capacity, multipy by the energy input and divide by the mass of air (or volume and density...), your answer won't be too far off for the first hour or two. After that, though, it will start to diverge quickly from the reality as other factors such as the heat absorbed by the objects in the room and the walls...and heat transfer through the walls becomes significant.

 

[mfb]

I have worked in rooms that size, with 2 kW heating. If they would have raised the temperature to 100 °C within one and a half hour, I would not have worked in those rooms!

 

[russ_watters]

I have worked in rooms that size, with 2 kW heating. If they would have raised the temperature to 100 °C within one and a half hour, I would not have worked in those rooms!

One of us did the calculation wrong because that's not the answer I got...

 

[CWatters]

While you put energy into the room some is escaping through the walls and via ventilation. Depending on the temperature inside and out and how well insulated it is, it might take more than 2kW to maintain a temperature let alone increase it.

The static temperature case (power in = power out) is usually easy to work out if you have data on the thermal properties of the walls, windows and ventilation. The dynamic case isn't quite so easy.

 

[mfb]

One of us did the calculation wrong because that's not the answer I got...

My calculation is linked in post #2.

 

[sophiecentaur]

Your calculation needs to include area of surfaces, rate of air change, distribution of 'warm' air in the room (what actually 'feels warm'). There are many links which discuss the practicalities involved (from the point of view of Heating Engineers. If you want a realistic answer, it may be worth reading a few as it's easy to miss factors out if you are trying to start from scratch and just use 'theory'.

 

[russ_watters]

My calculation is linked in post #2.

Whoops, I couldn't understand the funky units WA used, so I had to re-create it manually; I messed-up the time conversion and was low by a factor of 60.

The result is more surprising to me than it should be, but in any case means the divergence from the simple case (when heat transfer and heat capacity of objects in the room starts to matter) happens in just a few min.

 

[mfb]

Right. This is easy to test during the winter: open a window => it gets cold within minutes. Close it => ten minutes later, it is warm again (if you don't use a thermometer, leave the room as slow gradients are hard to notice). The short air exchange doesn't cool down the walls and other things, so once the window is closed they quickly re-heat the air in the room.

 

[baracoda]

Thank you guys for the responses!

What do you think of the following way of thinking:

1. I can calculate the amount of energy necessary to heat up the air:
[tex]W=mc_{air}\Delta T[/tex]
mass of the air enclosed in the room:
[tex]m=\rho V[/tex]
[tex]m=6*6*3*1.2=216\left [ kg \right ][/tex]
specific heat capacity:
[tex]c_{air}=1005\frac{J}{kgK}[/tex]
temperature difference (lets say now it is 20 Celsius degrees in the room and I want to heat up to 23):
[tex]\Delta T=3K[/tex]
So summing this up we have:
[tex]W=216kg*1005\frac{J}{kgK}*3K=651240J=0.1809kWh[/tex]
I have a heating system with power 2kW so in order to get 23 Celsius degrees I need to heat it for the following time:
[tex]\frac{W\left [ kWh \right ]}{P\left \[ kW \right \]}=\frac{0.1809}{2}=0.09\left [ h \right ]=5.4\left [ min \right ][/tex]

What if I somehow use it for calculating how everything will change in 1h? Is the solution ok?

P.S. 2kW heater was just an example.

 

[CWatters]

Is this a real world problem or just an exercise?

Real houses are ventilated and leak air quite badly. Some ventilation is also required by local regulations. This can be >0.6 air changes an hour. Quite a % error if not taken into account. Losses through walls and windows can be equally significant.

 

[mfb]

So summing this up we have:
[tex]W=216kg*1005\frac{J}{kgK}*3K=651240J=0.1809kWh[/tex]
I have a heating system with power 2kW so in order to get 23 Celsius degrees I need to heat it for the following time:
[tex]\frac{W\left [ kWh \right ]}{P\left[ kW \right]}=\frac{0.1809}{2}=0.09\left [ h \right ]=5.4\left [ min \right ][/tex]

What if I somehow use it for calculating how everything will change in 1h? Is the solution ok?

P.S. 2kW heater was just an example.

That's the calculation I linked to in post #2. It works for very short timescales (a few minutes) only, afterwards you have to take into account that the rest of the room (with much more mass than the air) will heat up as well.
By the way, the density of air is not 1kg/m³.

 

[baracoda]

Is this a real world problem or just an exercise?

No it is just an excercise. I am a programmer and I do not know physics. I am trying to figure out how to calculate how temperature will change in room during the hour (or other time span) and I want to skip the advanced parts of this problem, the easier for me to understand the better :)

That's the calculation I linked to in post #2

Oh.. you are right. Sorry.I have came up with such a model and I have given some parameters the default values . I am looking for a way to be able to calculate how the temperature will change in blue room in 1 hour if I will switch on the radiator or switch it off (or leave in the current state). Let's say the room is adjecent with 2 other rooms which temperature is 22 Celsius degrees. In the room there are also bed, wardrobe and desk with the dimensions given. Can you please help me to combine the @mfb solution with the situation below? Please do not take into consideration very sophisticated physical phenomena - it can be kind of controlled situation.

 

[russ_watters]

So, i'll repeat what the others said, in a different way:
In HVAC, the transient case you are investigating is usually totally ignored, because it has little real value. The steady-state case (balanced input and output heat) is what matters.

So can you explain what value the answer to your question will provide to you? What are you going to use this for?

 

[baracoda]

So, i'll repeat what the others said, in a different way:
In HVAC, the transient case you are investigating is usually totally ignored, because it has little real value. The steady-state case (balanced input and output heat) is what matters.

If I understand correctly, those changes I am trying to calculate are more or less right in situation: it is very cold in the room, I am switching on the radiator, and after one hour the temperature is x. Then those changes are not so big, and the other things in the room, together with walls etc play bigger role? Is that right? So is there any way to somehow incorporate the data I have put int the picture to calculate those temprature changes? I don't expect the exact perfect result, just more or less know how it behaves.

So can you explain what value the answer to your question will provide to you? What are you going to use this for?

Sure, I am trying to write sth like a simmulator - an application which uses machine learning algorithms for my classes. I thought that a heating simmulator would be a good idea. So one part of the task is just to calculate how the temperature will change after one hour. It should be a very simple case as it is not the main part of the project.

 

[mfb]

It is really hard to get a realistic model purely with theoretical predictions. Putting an actual heat source is a much better way to estimate how a room will react as there are so many factors that influence the temperature changes significantly.

 

[baracoda]

It is really hard to get a realistic model purely with theoretical predictions.

I see... there are many factors that influence the result...

Putting an actual heat source is a much better way to estimate how a room will react as there are so many factors that influence the temperature changes significantly.

You mean, that using the solution you have proposed in the first post would be the best idea? If yes can you please advise me how many Wats approximately the radiator should have to be near to realistic solution? And is there also some kind of easy way to calculate how much the temperature will drop if I will switch off this radiator?

 

[mfb]

You mean, that using the solution you have proposed in the first post would be the best idea?

No, that approach leads to answers that are completely wrong.

If yes can you please advise me how many Wats approximately the radiator should have to be near to realistic solution? And is there also some kind of easy way to calculate how much the temperature will drop if I will switch off this radiator?

I just said in my previous post that such an estimate is really problematic.

 

[baracoda]

I didn't realized that it is so problematic. Well, I think I will have to give up as there is no even very simplified way to deal with it. Anyway, thank you guys for your time.

 

[russ_watters]

I didn't realized that it is so problematic. Well, I think I will have to give up as there is no even very simplified way to deal with it. Anyway, thank you guys for your time.

Hang on: we didn't say you have to give up, we just said your approach won't work. Give me an hour; i'll show you how I would model this in a way conducive to programming. The key is to start with the steady state (with assumptions if necessary) and go backwards from there.

 

[russ_watters]

The way to go about it is to start with the steady-state and work backwards to the profile over time. In some ways, this can be almost as simple as the method you are taking, especially since you are doing programming, which makes numerical methods easy to implement.

Let's say you start with your 2kW heater and your 6x6x3m room and assume that that heater can hold that room stable at 20C in 0C weather. The heat flow out of the room at any temperature is a simple linear function of temperature (100W / 1C). What makes the room not go from 0 to 20C when you turn on the heater is the mass of the air, walls and furniture absorbing heat along the way.

Now, the physicists here would probably set up/derive an equation for that (I might be able to too if someone twisted my arm), but as a lazy engineer, I prefer an iterative/numerical method, with plug-in constants that I can adjust manually to model real-world performance.

So let's assume that this room of ours reaches 19C in one hour. Here's what that looks like:

 

[baracoda]

@russ_watters thank you very much for this example, it is much more clear to me than those complcated equations. Using ths method I can adjust the data as I want and also use it to predict how the temperature will drop if I will turn the heater off (I will have to look at the plot the other way as I think).

I have some questions:
1. By 0C weather you mean temperature outside?
2. What is the differenece between temperature and room heat?
3. What doest it mean that available heat is -20?
4. If I understand correctly the input data is in columns Temp, Room Heat and Heat Flux. The rest you calculated using some formulas. Is it possible that you would send me the excel file with this example as I have a problem with figuring out where did the values came from.
5. I am also not sure what is Room K and Room C.

Thank you.

 

[russ_watters]

@russ_watters thank you very much for this example, it is much more clear to me than those complcated equations. Using ths method I can adjust the data as I want and also use it to predict how the temperature will drop if I will turn the heater off (I will have to look at the plot the other way as I think).

You're welcome. FYI, the basis of this is Newton's Law of Cooling:
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

So yes, it works for both heating and cooling. Reading the full explanation of how it works may answer some of your questions, but...

1. By 0C weather you mean temperature outside?

Yes.

2. What is the differenece between temperature and room heat?

So, first a little mea culpa. "Room Heat" and "Available Heat" were somewhat redundant, or, rather, "Room Heat" was actually meaningless. I'll explain in more detail below, but here's a corrected spreadsheet, showing only "available heat" (notice the result is pretty much the same -- the "Room Heat" column wasn't really doing anything useful).

Newton's law of cooling is like emptying a bucket of water, except that instead of water, you have a container full of heat (thermal energy). "Available Heat" is the amount of heat that has to be added to the room in order to bring the temperature up to the desired 20C. "Room Heat" was intended to be the increment of change when it gets 1C warmer, but it was unnecessary and I didn't do it right anyway, so you can ignore that (and I've removed it). Temperature is just the temperature of the room at each time.

However:

3. What doest it mean that available heat is -20?

This is somewhat moot now, however please note that an iterative method like this is not exact. It has errors. Namely, it assumes a constant heat flux for an each 1C step, which isn't true - the reality is that the heat flux changes continuously. With the new version of the spreadsheet, the error is hidden, but note that in the true version of Newton's Law of Cooling you don't ever get exactly to your target temperature: the last line of the spreadsheet is always very significantly wrong. In the rest of the steps, the errors are relatively small.

4. If I understand correctly the input data is in columns Temp, Room Heat and Heat Flux. The rest you calculated using some formulas. Is it possible that you would send me the excel file with this example as I have a problem with figuring out where did the values came from.

Attached.

5. I am also not sure what is Room K and Room C.

They are my versions of constants that go into Newton's Law of cooling. K is the heat flow rate versus temperature difference: how well insulated the walls of the room are. C is the thermal capacity of the room: how much heat the interior spaces absorb. I believe in Newton's actual Law of Cooling they are combined into one constant.

 

[baracoda]

@russ_watters that is great! Thank you once again. I am studying it know and trying to understand and apply it to more real life example. I have found an information that for 30m2 room with 200kg of furnitures thermal capacity is more or less 10^7 J/K so I have to find appropriate room K, because now the results are completely wrong. Now I am trying to get the following:
- I have room temperature
- I have outdoor temperature
- I have max temperature in the room the heater can produce
- I am calculating how much time it will take to get to:
a) current room temperature - 2C
b) current room temperature + 2C
I am having a problem with this due to the following reasons:

Two questions as for know:
1. room K and room C use Kelvins and the temperatures provided are in Celsius. Is it ok?
2. I have a problem finding any information about heat flow rate value as I wanted to find estimated for my 6x6x3 room. How did you came up with the 200 W/K?

 

[CWatters]

furnitures thermal capacity is more or less 10^7 J/K

1. room K and room C use Kelvins and the temperatures provided are in Celsius. Is it ok?

J/K in English is "Joules per Kelvin". Absolute values of Centigrade and Kelvin are not the same but a 10C change in temperature is the same as a 10K change in temperature.

2. I have a problem finding any information about heat flow rate value as I wanted to find estimated for my 6x6x3 room. How did you came up with the 200 W/K?

This depends on how well insulated the room is. Look up Thermal Conductivity. This relates the heat flow to the thermal gradient across the wall. Normally a wall comprises layers of different materials each of which has a different thermal conductivity and a different thickness. It's possible to calculate the thermal conductivity of the overall sandwich.

For a wall it might be anywhere from 2 W/m²k for a solid brick/stone wall down to 0.1 W/m²k for an insulated wall. The units W/m²k are Watts per square meter per Kelvin.

For windows... a single-glazed window might have a U-value of 5.0W/m²K, double-glazed window perhaps 1.2-1.5 W/m²k and a triple-glazed under 1.0W/m²k.

So if your room has 6x6x3 walls (72 square meters) with a thermal conductivity of 0.5 and it's 20C inside and 5C outside the heat loss through the walls (only) would be..

Power (W) = 72 * 0.5 * (20-5) = 540W

but it could range from say 100W to 2000 W depending on how well insulated the walls are.

 

[CWatters]

PS Some insulation suppliers like Celotex have a U-Value (=thermal conductivity) calculator on their website which will work out the thermal conductivity of a wall or roof made of different layers.

 

[russ_watters]

The above responses were good, but to clarify; since no information was given about the room's construction, the constants I used were just totally made-up based on how a real room might perform. But there can be wide variations depending on the insulation, windows, sun, wind, infiltration (how airtight it is), etc.

 

[baracoda]

J/K in English is "Joules per Kelvin". Absolute values of Centigrade and Kelvin are not the same but a 10C change in temperature is the same as a 10K change in temperature.

Oh ok, I see the point.

Power (W) = 72 * 0.5 * (20-5) = 540W

Ok, I found some U-values and calculated that for my room Power (W) = 270 * (temp_inside - temp_outside), so for 20C inside and 5C outside it is 4050W.
How can I go from 4050W to the W/K unit? Or is it already 4050W/J?I have went through Newtons Law of Cooling and I understand the concept now (thank you for the link) but I am not sure where did you get the formulas to calculate available heat, heat flux and time interval from. Are those well known equations?

And how have you came up with thermal capacity of the room equal 4500 J/K?
Is it from the equation: E=m*c*deltla T
where:
m - mass in kg
c - specific heat capacity of the material
delta T - temperature change

I have assumed that the room has two internal walls made from concrete, two outer walls made of brick, one window and roof made of wood. I calculated its mass and found specific heat capacity and I have calculated that my room has heat capacity ~ 3 000 000 J. Is it also J/K?

Putting those calculated by me room K and room C in the excel I get a result that a room from 5 to 20 degrees will have to be heated for 16h. Am I doing sth wrong?

Have you used for something information about my imaginary 2000W heater?

 

[russ_watters]

Busy couple of days...

How can I go from 4050W to the W/K unit? Or is it already 4050W/J?
[snip]
Have you used for something information about my imaginary 2000W heater?

So, there's another issue with my spreadsheet. Somehow in playing around I got away from your scenario a bit. The initial heat flux is determined by the heater and, alternately, the room construction (insulation). The goal of this scenario is to start in one equilibrium, turn on a heater, and get to another equilibrium. I probably initially selected a 10K (or C -- I probably shouldn't have mixed them) temperature difference and then changed my mind and made it 20K. But the way I designed the spreadsheet, the temperature difference drives the heater output. So the starting heat flux is 4000 W with a 20K delta-T. You can change to a 10K delta-T to get 2000W or leave the delta-T and change the room K value to 100 W/K.

The "Heat Flux" is the net heat transfer into the room: the 2000W (or 4000W) of the heater, minus the heat loss to the walls. So if we start off in equilibrium at 0C, the heat loss is zero.

And how have you came up with thermal capacity of the room equal 4500 J/K?
Is it from the equation: E=m*c*deltla T
where:
m - mass in kg
c - specific heat capacity of the material
delta T - temperature change

I generated it manually - just plugging in numbers, not by trying to design a room. Newton's law of cooling let's you work backwards from the performace of a room to find the constants instead of trying to calculate the constants from the design of the room (since I didn't know anything about the design of the room).

I have assumed that the room has two internal walls made from concrete, two outer walls made of brick, one window and roof made of wood. I calculated its mass and found specific heat capacity and I have calculated that my room has heat capacity ~ 3 000 000 J. Is it also J/K?

No, heat capacity in Joules would be the total heat it can hold with a certain temperature difference. But there's a problem with your calculation: the room boundaries are not sharp. The walls are part of the boundary, so only some of the wall gets included in the heat capacity of the room. That's a lot of what makes your method so difficult and best ignored if you're just tyring to do some programming.

 

[CWatters]

How can I go from 4050W to the W/K unit? Or is it already 4050W/J?

4050W is the power needed to maintain a15C difference between inside and out. If you want that in W/K (or W/C) just divide by 15.

 

[baracoda]

Guys thank you very much for all your help!