Steady State Heat Flow Calculations

[vwalla]

To start, I am not an engineer. I am a programmer trying to write a steady state heat flow program for refractory linings for my family business.

I keep running into problems. With the heat flow programs I have seen, not one will ask to guess the coldface temps & the mean temps. Essentially it asks for material (k-factors, which I have), internal temp, velocity (fps), ambient temp, insulation thickness, & using an emissivity of .95 (typ.)

I have been studying North American Mfg. company's SSH transfer calculations & it involves looking up heat loss in a chart, & guessing the coldface temp & mean temps. Are there assumptions that I can make based on certain factors so my customes will not have to do this?

My program would be able to iterate the functions until sufficiently refined.

Any help would be appreciated.

 

[sailoday28]

As a programer, you should be able to step up the finite difference equations that govern heat conduction within the refactory. (Assuming the refactory is solid) This should include the thermal conductivity as a function of temperature and possibly direction.

The finite difference equations then would be limited by the appropriate physical geometry of the refactory.

The big problem other than geometry are the boundary conditions. Radiation (impacted by geometry of the refactory) and surroundings and local convection coefficients/ and or contact surfaces with other bodies via heat conduction.
Put these all together and you have solved your problem.
Of course many simplifications might be made dependent upon what the model is, ie, how thick compared to length, etc.

 

[corus]

You should be able to calculate an analytical solution of the equations by just taking T=aX+b and allowing k1.dT/dx=k2.dT/dx between materials. For refractory just assume a constant value of conductivity. Manufacturers rarely give values varying with temperature anyway. At the surfaces you could estimate a value of heat transfer coefficient to the ambient temperature. If the surface is next to a liquid it would be reasonable to assume that the surface was at the liquid bulk temperarture. For heat loss to air take a heat transfer coefficient value of about 12W/m^2 K to an ambient 20C. At the surface your equation would be -k.a=h.(T-Ta). You need to solve for a and b.

corus

 

[vwalla]

Thanks to both of you. I will let you know how things turn out.

 

[davefitz]

It may be a little more complicated than prior messages indicate.

For many canned programs , and for many applications, the outer surface may be lagged with aluminum lagging ( e=0.10, not 0.95). Also, in many applicaitons, all accesible outer surfaces may be requried to be below 140F for OSHA safety reasons, or else a wire barrier be added to prevent injury.

Inside the furnace or process, the inside emissivity ( refractory to process) may be a complicated function of inside surface temp, process gas temp, and surface deposits.

And finally, the outside heat transfer coefficient can vary due to wind effects or due to natural convection ( if it is a tall structure).

If you need the 0.95 outer emissivity , then use a painted surface , or a bare steel surface, or anodized aluminum lagging.

 

[vwalla]

I am currently working with our material supplier on this & they have given me a list of coefficients to use for the material & I am using a Heat Flow program from another supplier. I can email this program (excel document) if anyone would like to take a look at what I am trying to do. Also I can email the North American Mfg procedures.

I can alos post the programs on our website if it would be easier.

With the heat flow program, you can see that cold face is not guessed, but with the North American Mfg procedures, it asks you to gues the coldface.

I have copied our coeff. into this program, with less than desirable results.

Are there different ways to find the coeff?

 

[IRstuff]

Some are "relatively" easy to measure, others less so.

> Thermal conductivity -- usually reasonably reliable from supplier

> Emissivity -- probably a book value, but can be better quantified with a radiometer

> Environmental irradiance -- application specific, requiring in-situ radiometric measurements

> Convective coefficient -- application specific, requiring in-situ radiometric measurement

> Interface thermal conductivity between materials -- extremely difficult to determine, and may not matter. Requires some in-situ measurements to determine significance.

You can send to arbiter007 at hotmail.com


TTFN

 

[TangoCleveland]

vwalla,

Track down a copy of ASTM C680, "Standard Practice for
Estimate of the Heat Gain or Loss and the Surface
Temperatures of Insulated Flat, Cylindrical, and Spherical
Systems by Use of Computer Programs". It's a useful document for the kind of calculation you're doing. Particularly handy are the calculations for free and forced convection in air, and for air properties. Also includes how to handle emissivity.

I'd like to look at your spreadsheet. Please email to gmucs@netzero.com

Larry

 

[vwalla]

Larry-

ASTM C680....well worth buying. Thanks & let me know your thoughts on the spread sheet.

IRstuff-

Let me know your thoughts also.

Thanks fellas!

 

[IRstuff]

Thanks for the email.

The NA worksheet uses the guess values to start the iterative process, due to the fact that the materials have thermal conductivities that vary with temperature.

As discussed in the document, the initial guesses are used to select thermal conductivities which are then used to determine the intermediate temperatures. This first calculation is then used to refine the thermal conductivities, etc., until the new intermediate temperature are essentially unchanged from the previous set.

This could be mechanized on a spreadsheet in a similar fashion, using Solver or Goal-Seek routines.

In thermal analysis programs, the process does not require guess values because they can be estimated based on the approximate k-values and thicknesses of the materials, i.e., if all the materials were treated as the same material with different thicknesses, you could quickly determine guess values for the intermediate temperatures. This is the basic premise for the calculations used in Step C of the worksheet, where the temperature is treated as a voltage across 3 resistors in series.

TTFN

 

[sailoday28]

From the previous responses, I take it that the problem is one dimensional.

 

[vwalla]

I am not sure what you mean by one dimensional.

All of our heat flows that we do are basically taking a wall of infinte height & depth & calculating the cold face temp by knowing the inside temp, k-factor, insulation thickness & if it is a sidewall, hearth, or cylinder.

Would that be considered one dimensional?

Thanks

 

[IRstuff]

One-dimensional, since all you seem to care about is thickness, and since you indicate infinite height and depth

TTFN

 

[chicopee]

vwalla-The problem w/ heat transfer equations involving radiation is that you deal w/ temperatures to the fourth power,so in order to solve such equation you will be required to make an initial estimate of the surface temperature on which radiation is impinging. W/ this initial estimate, you would calculate heat transfer from radiant heat and convection on the hot side. This combine value will equal heat transfer by conduction from which you can calculate the cold side surface temperature. Heat transfer from the cold side to ambient will be primarily thru convection and perhaps involving could involved heat transfer by radiation depending on the surface temperature of the cold side. Remembering that the heat transfer on the hot side must equal heat transfer by conduction which in turn equals convective(and perhaps radiative)heat loss on the cold side, you may have to hone in on these surface temperature estimates until all these rate of heat transfer equal to each other.
I read your question several time and I was not sure what you were trying to say. I eventually surmise that you were asking about making initial estimates of the surface temperature to make this program work.

 

[IRstuff]

All the conductivity values are temperature dependent, hence, you need to start with a guess for the average temperature to determine the conductivity to determine the amount of heat lost

TTFN

 

[chicopee]

I agree w/ temperature dependent conductivity values; however,conductivity values given in heat transfer books are for temperature ranges and consequently adjustements can be made if the material temperature starts shifting from that range.