Heat generated in an enclosure?

[damightytom]

Hi I have an enclosure with an electrical heater that runs on 5W.

Im trying to figuring out how warm the enclosure gets after a set amount of time.
I need help with the strategy on how to tacle this problem.

The heater warms up the air inside the enclosure (Specific heat) then the enclosure absorbs some of that heat and then radiate that to the outside depending on the sun radiation and ambient temperature.

I have been stuck on this for 2 days and I really could use some help confirming or helping me along the way on figuring out what I need to solve it.

 

[billy_joule]

What have you got so far? Why not assess the steady state before the much harder transient? The result might tell you enough that you don't need to consider transient

 

[damightytom]

Im trying to compare two enclosures, one is plastic and one is aluminium.
So I figured I would calculate the air temperature inside the enclosure.

For the plastic enclosure with an volume of 0,00097m³ (to simplify I make the assumption that the enclosure is empty and filled with air. That has a Specifik heat of 716 Kj / Kg and air with a density of 1,3kg/m³ the air inside the enclosure would weigh 0,00126kg.

So Watts / Specifik heat /weight would give me a temperature.
5 / 716 / 0,00126 = 5,55 C°
(I guess I can add time to the Watts (seconds) but I want to start at t(1) for starters to make sure I get the basics done)
In my mind that's the temperature rise from ambient temperature that will transfer into the enclosure.
What I'm looking into now is calculating the heat transfer rate of the specifik plastic and aluminium to determine how much of that generated heat is lost through the enclosure.

In the end I want to compare the temperature for a given time in both enclosures to see if i raises over any temperatures that might be damaging to the electronics.

 

[billy_joule]

Your unit for specific heat is not right.
Are you trying to use Q=m Cp deltaT? Your units are not making sense.

It's still not clear why you don't want to consider steady state.

 

[damightytom]

This is a new field for me so its a struggle.
Care to explain steady state and how I could use it?

deltaT = Q/ (mass*CP)
deltaT = (5w*t) / ((density*volume)*CP)

Am I getting that right or am I still missunderstanding it?

 

[damightytom]

Your unit for specific heat is not right.
Are you trying to use Q=m Cp deltaT? Your units are not making sense.

It's still not clear why you don't want to consider steady state.

Thank you. I see now that now. I use a nominal value of specific heat now on air witch is around 1.0 KJ/Kg.K

With a volume of 0,0026m^3 and a weight of 0.00338kg of air in that volume I get a temperature rise of 1,5 Kelvin from applying 5W.
Does that sound reasonable?

And now when I want to calculate the heat dissipation from the enclosure material on the "outside" can i retract that from the temperature on the inside of the enclosure?

Sorry I feel really dumb too, but i appreciate the help i can get.

 

[russ_watters]

With a volume of 0,0026m^3 and a weight of 0.00338kg of air in that volume I get a temperature rise of 1,5 Kelvin from applying 5W.
Does that sound reasonable?

No. Watts has units of time, so your temperature rise must be over a period of time as well.

And now when I want to calculate the heat dissipation from the enclosure material on the "outside" can i retract that from the temperature on the inside of the enclosure?

Sorry I feel really dumb too, but i appreciate the help i can get.

It's ok, but you are approaching this question backwards. In steady-state, heat input and output are the same, so the temperature in the enclosure is determined exclusively by how effectively heat is transferred out of the enclosure. So you need to focus on the details of the enclosure and how it transfers heat out: How big is it? What is it made of? Does it have a fan to exchange air with the outside?

 

[damightytom]

No. Watts has units of time, so your temperature rise must be over a period of time as well.

It's ok, but you are approaching this question backwards. In steady-state, heat input and output are the same, so the temperature in the enclosure is determined exclusively by how effectively heat is transferred out of the enclosure. So you need to focus on the details of the enclosure and how it transfers heat out: How big is it? What is it made of? Does it have a fan to exchange air with the outside?

Thank you!

Alright I have the heat transfer coefficiants for Aluminium = 5,882 W/m^2 and Some plastic that's similar to the one I am planning to use = 3,448W/m^2
It does not have any fans it's a fully enclosed system. It's an IP-classified electronic box.

The aluminium one is a cylinder shape R=56,5mm Lenght=250mm thickness=3.0mm inner surface area = 111000 m^2
The plastic one is not uniform but generally its 92x106x150 with a thickness of 3,0mm. inner surface area = 355680mm^2

 

[damightytom]

I'm having trouble wrapping my head around the Heat flow rate. If the outside temperature is 20 °C the temperature inside the enclosure is very likely to be higher. Which results in a Delta Temperature of a positive inside the enclosure. The heat "wants" to go out to the coocler ambient temperature.

If I know the outside temperature, the energy output in the enclosure from the electronics. how can I come to the conclusion of inside temperature?

 

[damightytom]

Really could use some guidance on how I can calculate the Heat flow rate without knowing the temperature inside the enclosure.

 

[billy_joule]

at steady state the outside surface transfers 5W to the surroundings regardless of what the box is made off.
The surface temp can be calculated from that.
You then work inwards from the surface temp to find the inner temps.

These should get you started:
http://www.sfu.ca/~mbahrami/ENSC 388/Notes/Staedy Conduction Heat Transfer.pdf
http://www.mhhe.com/engcs/mech/holman/graphics/samplech_3.pdf

 

[russ_watters]

I can provide more help later, this just requires more than a blurb from a cell phone and I was busy last night.

 

[russ_watters]

Ok...

Im trying to compare two enclosures, one is plastic and one is aluminium.
So I figured I would calculate the air temperature inside the enclosure.

For the plastic enclosure with an volume of 0,00097m³ (to simplify I make the assumption that the enclosure is empty and filled with air. That has a Specifik heat of 716 Kj / Kg and air with a density of 1,3kg/m³ the air inside the enclosure would weigh 0,00126kg.

So Watts / Specifik heat /weight would give me a temperature.
5 / 716 / 0,00126 = 5,55 C°
(I guess I can add time to the Watts (seconds) but I want to start at t(1) for starters to make sure I get the basics done)
In my mind that's the temperature rise from ambient temperature that will transfer into the enclosure.
What I'm looking into now is calculating the heat transfer rate of the specifik plastic and aluminium to determine how much of that generated heat is lost through the enclosure.

In the end I want to compare the temperature for a given time in both enclosures to see if i raises over any temperatures that might be damaging to the electronics.

Heat capacity inside the enclosure doesn't matter because at steady state it isn't gaining or losing any heat (net). All that matters for is for how long it takes to get to the steady state.

What matters is how well insulated it is.

So, plastic isn't that easy to find insulation values for, but perhaps it is similar to vinyl siding -- it's probably better.
http://inspectapedia.com/insulation/Insulation-Values-Table.htm
In American units, that's 0.61, or in SI units, 0.107 k-m^2/W. From here, it's a resistance problem:
DT = P*R/A

DT = temperature differene
P = Power
A = Surface area

This will get you in the ballpark.

 

[damightytom]

Ok...
Heat capacity inside the enclosure doesn't matter because at steady state it isn't gaining or losing any heat (net). All that matters for is for how long it takes to get to the steady state.

What matters is how well insulated it is.

So, plastic isn't that easy to find insulation values for, but perhaps it is similar to vinyl siding -- it's probably better.
http://inspectapedia.com/insulation/Insulation-Values-Table.htm
In American units, that's 0.61, or in SI units, 0.107 k-m^2/W. From here, it's a resistance problem:
DT = P*R/A

DT = temperature differene
P = Power
A = Surface area

This will get you in the ballpark.

I have been ill for the last couple of days so haven't been at work, but I'm ready to tackle this problem again.
I'll study your link and the ones above yours and I'll will probably come back with questions if I don't understand it.

Thank you for the assistance.

 

[damightytom]

Ok...
Heat capacity inside the enclosure doesn't matter because at steady state it isn't gaining or losing any heat (net). All that matters for is for how long it takes to get to the steady state.

What matters is how well insulated it is.

So, plastic isn't that easy to find insulation values for, but perhaps it is similar to vinyl siding -- it's probably better.
http://inspectapedia.com/insulation/Insulation-Values-Table.htm
In American units, that's 0.61, or in SI units, 0.107 k-m^2/W. From here, it's a resistance problem:
DT = P*R/A

DT = temperature differene
P = Power
A = Surface area

This will get you in the ballpark.

Hi again, yes this really got me into the ballpark. I helped me a lot to understand my problem, at the time you answered i don't think i really understood what happened.

since i have the U-values for both materials and their thickness, 3mm i can calculate R with the U= 1/R formula.

U aluminium = 5,882
U plastic = 3,448

R alu = 1/5,882 = 0,17
R plastic = 1/3,448 = 0,29

P = 5W

Area alu = 0,111m^2
Area plastic = 0,067565m^2

DT aluminum = (5*0,17)/0,111 = 0,094 K
DT plastic = (5*0,29)/0,06756 = 0,098 K

These values seems extremely low.
how do you see it?

 

[thankz]

maybe this will help, I generally don't do the math for other people.

 

[damightytom]

maybe this will help, I generally don't do the math for other people.

Thanks a lot for that picture.
Unfortunately your kind efforts have been in vain. I don't understand how that picture can help me.

 

[damightytom]

Hello.
I was wondering if anyone could give me their perspective on my problem.
I know it's a bit tabu doing the maths and calculations for other people but I am at a loss here. I lack the textbooks and therefore the basics i need to understand what you guys are trying to tell me.

 

[Rive]

Hi,

What I would do: take the surface of the box as a heatsink. You can find some clues about calculating a thermal resistance of some kind of heatsinks/surfaces: also, you can combine the resistances and calculate the resistance of the box. By using the definition of 'thermal resistance' you might get the difference of the ambient internal and external temperature.

 

[thermal1]

Hi,
You need to calculate the internal, enclosure wall and external thermal resistances of the enclosure sum these values and multiply it by the amount of heat dissipated by the internal heater. These resistances are determined by the enclosure dimensions, enclosure material thermal conductivity and surface emissivity. The calculations are too lengthy and detailed to explain here. Check out the article below that provides the detail. Keep in mind that this is for steady state only.
[PLAIN]http://www.heatsinkcalculato...e-the-temperature-rise-in-a-sealed-enclosure/[/PLAIN]
http://www.heatsinkcalculator.com/blog/how-to-calculate-the-temperature-rise-in-a-sealed-enclosure/

 

[CWatters]

Presumably you can't just put a thermocouple inside an measure the temperature?

 

[Tom.G]

Let's try this approach.
'R' value for a metal box= 1 BTU/Hr/sq.ft./degree F (for the surfaces exposed to free air)
1 BTU= 3.41Watts
5Watts= 1.47BTU

That should be enough to get you started.

It's going to be an awful lot hotter if it is in direct sun. For a quick, conservative, approximation, Solar insolation can be a little under 1Kw per square meter. (That's Noon in the desert in the Southwest U.S.) The amount absorbed by the box will be decreased by it's emissivity (0.8 - 0.9 as a guess) and by the cosine of the incident angle of the Sun on the surface.

 

[Babadag]

See also:
http://www.dii.unipd.it/~gobbo/dida...BB_swithgear_manual_E11/ABB_11_E_01_druck.pdf
section 1.2.5 Heat Transfer