Heat transfer basics

[zappedagain]

This thread is branched from

http://www.eng-tips.com/threadminder.cfm?pid=391

Nice thread! I'm looking for the static heat rise of a hand held enclosure, so let me see if I have this correct:

Tia = inside air temp
Tis = inside surface temperature
Tos = outside surface temperature
Tamb = ambient temperature

and

There is a heat transfer rate Q1 defined by the heaters
There is a heat transfer rate Q2 defined by the pair T1a---Tis
There is a heat transfer rate Q3 defined by the pair Tis----Tos
There is a heat transfer rate Q4 defined by the pair Tos---Tamb

The 3 equations:
Q1=Q2
Q2=Q3
Q3=Q4

Q1 - power dissipated in the enclosure

Q2 - temperature rise in the box (this has a positive relationship to the thermal resistivity of the object dissipating the power, correct?). Q2=mc delta T (heat = mass *specific heat constant of air * change in temperature), correct?

Q3 - If I ignore radiation, this is defined by the thermal conductivity of the enclosure, correct?

Q4 - How would I calculate this for a given volume, such as a 2"x2"x2" box? Is there some estimate I can use for a ballpark number?

Set Q1=Q2=Q3=Q4 and the result is a ballpark temperature for Tia, Tis, and Tos, correct?

As I'm ignoring radiation (it should lower the temperature as long as I don't have an external heat source, so this is worst case). Are there other parameters that I might want to include?

Thanks,

John D

 

[IRstuff]

Your link simply points to the forum, not a particular thread.

The simultaneous equations only apply for steady state. Q4 is usually the easiest to calculate; calculate the free convection surface area, and choose your best guess at the free convection coefficient, which ranges from 2.5 W/m^2-K to 8 W/m^2-K. In a steady state problem, the internal heat dissipation plus any absorbed heat must all appear on the surface of your box.

I would couch the problem slight differently, in that your 3 unknowns are the 3 temperatures other than the ambient, which you should already know. The heat flows for a steady state problem MUST be equal, since there is conservation of energy, and there are no invisible sinks or sources. The same 3 equations with 3 temperature unknowns.

TTFN

FAQ731-376

 

[someguy79]

Q2
It seems as if what's dissipating heat in your box has some mass flow associated. Check your units and make sure you are correctly using a relationship to get units of power or energy. mdot*Cp*(T1a-Tis) would only apply to a continuous flow at constant pressure where the fluid is goes from an average temperature of T1a to an average temperature of Tis. The units I've used here assume power is to be calculated, though I'm often sloppy in my description of heating power vs. heat.

Q3
The simplest equation you are likely to use for Q3 is for a solid of constant temperature Tis on the one face and Tos on the other.
Q3=kA(Tis-Tos)/L
L=thickness of solid
A=area of solid
k=thermal conductivity of solid material

Q4
This is more complicated. You will need to know what type of model to use for convection. In essence, you need to know whether you have forced or free (a.k.a. natural) convection. Then you'll have to apply the proper boundary conditions and figure out the convection coefficients. Then you can apply the equation(s) for convection.
Q4=hA(Tos-Tamb)
h=convection coefficient

You mentioned that this is a hand-held object. If it is truly hand-held, you may have an external heat source, the hand. It may work to assume the hand is a constant temperature object in contact with the box surface.

 

[zappedagain]

Oops. here is the link to tunalover's thread:

thread391-280084

Q2 may be the wrong equation then. My heat source is some electronic gizmo, so the power dissipation ia easy to calculate. I'm ignoring conduction on the cabling (again, I'm getting worst case numbers). As my heat source is a stationary solid, the only motion within the enclosure will be convection currents. I assumed that I use the mass of the air volume within the enclosure (at a given altitude). Or should I be using a different equation?

Z

 

[IRstuff]

I glossed over your box dimensions the first time around. How thick are the walls? Your box is small enough that if the heat source is large relative to the interior volume of the box, the actual heat transfer would be pure conduction. For an air gap of less than about 1/2-in, the process is mostly conduction, and not convection. Even with a small source, the smallness of the box itself might argue for conduction instead of convection. However, if you are trying to find the worst case, convection would give you a higher internal temperature.

Is there a solar load?

TTFN

FAQ731-376

 

[zappedagain]

Interesting. I hadn't thought about the air gap size. My heat source is PCB mounted typically, so there isn't a direct conduction path except for in higher power applications. My initial thought was that conduction wouldn't be significant.

I can see where this starts to get tricky and your assumptions can quickly go astray. It looks like I have a bit more reading to do.

Typically the walls are fairly thin, 0.06"-0.18".

Z

 

[IRstuff]

In case I wasn't clear; I'm referring to thermal conduction through the air itself. This has to do with the thickness of the boundary layer. If the air gap is smaller than the boundary layer thickness, then there is little possibility of actual air circulation, since there's no pathway for the air to get to the bottom of the structure without colliding with the upwelling air.



TTFN

FAQ731-376

 

[zappedagain]

Ah, so I treat my thin air as a solid! What goes up possibly may not come down!

Z

 

[IRstuff]

But, like said, in the worst case, you'd go with convection, if that's the objective.

TTFN

FAQ731-376

 

[zappedagain]

Good point. If I want to take both conduction and convection into account, how do I account for their 'parallel' nature? Do I need to take a geometric mean, RMS sum, or linear mean to combine the effects?

Z

 

[IRstuff]

The conductivity drops substantially below the convection at a certain point. According to the Mathcad model in one of their old E-books, that point is around 13 mm, but I have no substantiation for that.

In any case, putting them in parallel may not make sense physically, since they are part and parcel, i.e., conduction is the direct genesis of convection. You conduct the heat into the air, and the increased buoyancy leads to the convection current.

So, they're actually in series, and the 13-mm transition point would be where the conduction produces sufficient buoyancy to start convection. Obviously, the actual physics is much more complicated, but that's the lumped model explanation, I think.

The bottom line is that you know the power dissipation of your electronics, and can work backwards from the external surface convection, i.e., for your little box, assuming 5 of the 6 surfaces are convecting by the same amount (a gross approximation), and say, 0.5W, we get about 15.5°C rise for htc.external=2.5W/m^2-K. If we assume roughly the same rise internally, you get an internal air temp of about 31°C above external ambient. That's neglecting the drop across the housing, which ought to be minimal.



TTFN

FAQ731-376

 

[electricpete]

I haven't studied the difference between all the scenario's, but the part quoted in original post comes from this thread:
thread391-280518

=====================================
(2B)+(2B)' ?

 

[zappedagain]

Thanks Pete! It does seem that I was link-challenged last week. :-(

John D