iterative model, conducted heat is a problem 2

[someguy79]

I'm trying to model a metal box sitting on the ground. It's 3" thick steel, painted white, and not insulated. It sits in the sun on a hot day with no wind and comes to steady state. Inside the box is a small heater.

The goal is to estimate the internal air temperature.

I've been trying to model the box by developing a heat balance on its external surface.

I know the solar radiation heat input (including incident angles, direct, reflected and diffuse components), as well as background radiation heat input (from ambient temperature, and absorbtivity).

Convection is assumed to be natural convection driven by difference between surface temperature and ambient temperature.

Net heat conducted (through all walls of box) heat is equal to the heater power inside the box.

The difficulty comes because I do not know surface temperature(s) of the box.

Solve this I've been trying to create an iterative model to get a solution. I'm finding that when I vastly increase thermal resistance to conduction (as if insulated) I can get the model to converge and give reasonable results. Unfortunately, even when I work the model to converge at smaller and smaller thermal resistances, at some point it fails and diverges.

I also tried looking at the box as a lumped capacitance (isothermal walls) but I end up with conduction equations that don't work.

Is there another strategy I should be looking at?

 

[prex]

From your words "Net heat conducted (through all walls of box) heat is equal to the heater power inside the box." I understand that you are dealing with a steady state situation?
In this case the external surface temperature of the box is simply and directly (not iteratively) determined by equating the heat flux (=heater power) to the total heat lost: convection, radiation to the surroundings, minus the heat input by radiation, plus, if relevant, the heat lost by conduction to the support of the box. All these quantities depend more or less linearly on box wall temperature, so the solution is straightforward.

prex

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[Twoballcane]

What does your hand calculations (or heat flow rate q=(1/R)deltaT) tell you?

Tobalcane
"If you avoid failure, you also avoid success."

 

[zekeman]

You have to solve the simple 3x3 linearized equation set with variables T1a, Tis, Tos.

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

T1a---Tis----Tos------Tamb

Think of these as nodes where the net heat to each node is zero; thus you write 3 equations, one for each node equated to zero.
So the T1a node gets the inside generator minus the convective exchange to Tia, the Tos node gets the solar input minus convective and radiative to ambient minus conduction to the Tis node, etc.

 

[electricpete]

The linearized approach can give a good starting value for a more exact linearized solution.

When you say "fail to converge" in an iterative approach, I wonder what kind of approach you are using.

How about a simple time domain simulation and let it run until it reaches steady state.

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

 

[someguy79]

Yes, it's a steady state problem.

Yes, I have the equations for conduction through the wall, convection to the ambient air, and radiation from the wall, from the sun and from the surroundings.

The major problem I have is that I do not have all the temperatures. If I did, it would be simple. According to Zekeman's problem setup, I only have outside ambient tempertaure (Tamb). I have to solve for the others. Obviously, you end up with more unknowns than equations.

If I can assume that the internal air is mixed well, then the internal air temperature (T1a) is very near to the internal wall temperture (Tis). It's not a perfect assumption, but it shouldn't be too bad.

To get around the problem of having more unknowns than equations, The external surface temperatures (Tos) are assumed. Then the heat flows are computed for radiation and convection. Through the heat balance on each surface the conduction heat is computed. From the conduction heat the internal temperature is computed. Then, the internal temperature (Tis) can be put back into a heat balance equations to back solve for external surface temperatures (Tos). These (Tos) are then used to drive the next iteration of the calc.

If it's done right, a solution will converge. The problem I have is that the conduction resistance is so low that it gets unstable and will not converge. If I add thermal resistance to the conduction path (as if it's insulated), it all seems to work just fine.

 

[electricpete]

The major problem I have is that I do not have all the temperatures. If I did, it would be simple. According to Zekeman's problem setup, I only have outside ambient tempertaure (Tamb). I have to solve for the others. Obviously, you end up with more unknowns than equations.

I think you have 3 indepdent equations.
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

You mentioned interative solution. If not time solution I imagine you are plugging in initial guess for T1a, Tis, Tos, Tamb and solving for new values which are plugged in as a new guess and repeat. You can try underrelaxation or overrelaxation to change the convergence of the model. I think underrelaxation tends to improve stability, overrelaxation tends to improve speed of convergence.

In an underrelaxed approach, you only move part way (some fraction alpha < 1) from the old solution to the new solution at each step. For example if T1 was 100 as initial guess and predicted as 101 at the end of the 1st iteration, you don't use 101 but 100+alpha * (101-100) where alpha <1 for underrelaxed and > 1 for overrelaxed


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

 

[electricpete]

Also radiative heating and/or cooling are included in Q4, but does not add any new variables.

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

 

[electricpete]

Correction in bold:

The linearized approach can give a good starting value for a more exact non-linear solution.

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

 

[electricpete]

in the form written above, I think the 3 equations can be solved sequentially (rather than simultaneously).

First solve f(Tos,Tamb)=Q for Tos
then solve f(Tis, Tos)=Q for Tis
last solve f(T1a, Tis)=Q for T1a


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

 

[someguy79]

Perhaps, I've been putting too many constraints on the model. the internal air temperature (T1a) is assumed to be the same as inside wall surface temperature (Tis). This might not be allowing for enough of a difference in temperature between the walls of the box. If conduction resistance is very low (think Biot number ~ 0.0001) it's like a lumped capacitance and should have constant temperature. In order for heat to balance, the different walls may need to have significantly different surface temperatures. These conditions are in conflict and it becomes difficult, if not impossible to get a solution out of the model.

I find it interesting that such a model has produced reasonable results with higher conduction resistances, but no other changes.

electricpete,

I hadn't heard of underrelaxation or overrelaxation before, but I have, in effect tried the former. The model will alternately increase then decrease a particular temperature being iterated. In effect they bounce around above and below what would appear to be the solution. In cases where these increases and decreases diminish, the solution converges. In cases where these grow, the solution diverges.

So, when I pick a value between the last iteration and the next computed value, The temperature tends to converge quicker.

 

[IRstuff]

"I've been putting too many constraints on the model. the internal air temperature (T1a) is assumed to be the same as inside wall surface temperature (Tis). "

>> Can't do that, by definition. No delta T, no heat transfer.

"The difficulty comes because I do not know surface temperature(s) of the box"

>> Not necessary. You have a set of simultaneous equations; the surface temperature of the box will come out of the solution to the equations. The only temperature you ostensibly know is the ambient temperature. I'm not sure how "Solve this I've been trying to create an iterative model to get a solution" Are you doing this by hand or using a solver routine in a program?

>> Posting your actual problem parameters and attempted solution might be useful.

TTFN

FAQ731-376

 

[someguy79]

I've been using Mathcad 14 to run the calculation. It's fairly large. Iteration is done by manual input.

The ambient is just a maximum daily high temperature for the installation location. (115°F)

If anyone really feels like going through this, here's a starting point.

I expect to see temperatures close to what you get for a car in the sun on a hot day. (150-200°F) Of course, this isn't a car, just a steel box painted white.

Q1 = 460W
Tamb = 115°F

no wind
box sits near the ground and is completely shaded.

insolation on top = 286Btu/h/sq-ft
insolation on bottom = 0 Btu/h/sq-ft
insolation on North side = 46 Btu/h/sq-ft
insolation on South side = 175 Btu/h/sq-ft
insolation on East side = 46 Btu/h/sq-ft
insolation on West side = 118 Btu/h/sq-ft
Insolation includes direct sunlight, diffuse sunlight, and reflected sunlight. Background radiation from ambient must be accounted for separately.

box material is 2" thick A36 plate

surface is white paint
emissivity assumed to be 0.9
graybody assumed

top area = bottom area = 11300 sq-in
North side area = South side area = 5640 sq-in
East side area = West side area = 2590 sq-in

 

[IRstuff]

>> If it's "completely shaded" why are there insolation values? Your 286 BTU/h/ft^2 = 902W/m^2. MIL-HDBK-310 lists maximum solar load at 1120 W/m^2, so that's almost full sunlight.

>> What's the absorptivity? You cannot use emissivity to calculate solar absorption, since the wavelengths being absorbed are not the same as the emitted wavelengths.

TTFN

FAQ731-376

 

[someguy79]

Sorry about that. The BOTTOM of the box is completely shaded.

Also, the wall thickness of all sides and top is 3".

 

[someguy79]

Assuming a greybody, absorbtivity is equal to emissivity. Though I must confess, the conditions where this is valid are a bit fuzzy to me.

The only values I could find on absorbtvity of a white paint (near 0.26) were so low as to push the surface temperature below the ambient temperature in some cases. In other words they made for nonsense results.

It's a conservative method to leave the absorptivity high. I'm concerned that the internal temperatures in the box will exceed rated temperatures for equipment therein.

 

[IRstuff]

If your temperatures came out goofy, it's not from the absorptivity. Physically, your box, in daytime, must be no lower than the ambient temperature. If it's not, then there's something wrong in your solve block.

With a 46°C air temperature, it wouldn't take much to get above industrial temp of 70°, or even 85°C.

TTFN

FAQ731-376

 

[someguy79]

If you run a case with no heat generation in the box
absorbtivity = alpha
emissivity = epsilon
alpha << epsilon
Boltzmann's constant = sigma
Q1 = 0 (for simplicity, different from case above)
E = solar insolation
A = area
h = convection coefficient

heat balances start to look like this:
alpha*E*A + alpha*sigma*A*Tamb^4 - epsilon*sigma*A*Tos^4 - h*A*(Tos-Tamb) = 0

If
Tos < Tamb
Then at some point
alpha*E*A + alpha*sigma*A*Tamb^4 < epsilon*sigma*A*Tos^4

This situation would not make sense intuitively. The system is below ambient temperature. Consequently, the low absorptivity values I found are not valid in this way. That's why I went to a much higher value. The best thing I can think of is the low absorptivity value was supposed to be associated with a particular spectrum.

 

[IRstuff]

Your second alpha should be an epsilon, since the ambient is roughly the same temperature as the box surface, so the wavelengths are roughly about the same. Alpha should only be applied to the insolation, since the sun's blackbody temperature is on the order of 6000K, whereas all terrestrial environments are no more than 373K.

Note also, that the radiative ambient is not necessarily the same as the air ambient, although they are usually, but not always the same. For example, clear sky has an apparent radiative temperature on the order of 270K, i.e., below freezing. For completeness, each side of your box would need to be evaluated for what it actually "sees" and that will determine what radiative background temperature to use.

The only time that heat balance equation goes negative is when sky temperature is very cold, say, 250 K, in which case, it's possible to get a surface temperature below the ambient temperature, because the heat is being radiated away faster than convection can warm it back up. If you have Mathcad, there is a file on the PTC resource center that has that particular calculation:

http://www.ptc.com/appserver/wcms/standards/textoimgothumb.jsp?&im_dbkey=59986&icg_dbkey=888

Generally, nonesensical solutions usually mean the solver had convergence problems, and usually, reformulating or rearranging the equations can clear up the nonconvergence.

TTFN

FAQ731-376

 

[prex]

I'm starting to get confused about what you are looking for: I understand that you need to estimate the temperature of the equipment inside the box.
Let me start again.
You have a box with massive carbon steel walls and you know the heat added inside and from solar radiation outside.
You can generally assume a heat exchange coefficient to ambient calm air of 10 W/sq m/deg C : this includes the effect of radiation (50% contribution roughly) and is not valid only when there is a large contribution of radiation to open sky (especially in the night).
Now you have all the data required to calculate the outer temperature of the box, that, owing to its more than abundant thickness, you can assume to be uniform. The calculation is direct, no iteration required.
Now you go backwards, as stated in your first post, and calculate the deltaT across the wall (if significant) and the deltaT on the inner wall face, using only the heater power as the transmitted heat.
So no iteration required.
A better explanation of which equations you used in Mathcad would help in understanding what you are doing.

prex

http://www.xcalcs.com

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