I want to design Furnace refractory lining. I know about inside temperature, outside desired temperature and general refractory lining parameters ( e.g, density of different linners, heat conductivity and thickness etc). I need to calculate temperature drop across different refractory layers. Is there any formulation?
Hi Mr. Sabyasachi,
Please refer the link below
I hope this is what you were looking for. In case you would require any further details, you can look into the book "Heat Transfer" by Viktor P. Isačenko or in fact any book on Heat Transfer.
Regards,
Shreyas Harsha
http://nptel.ac.in/courses/113104058/mme_pdf/Lecture29.pdf
Some materials do not behave isotropic manner in heat transfer. Still others, mainly ceramics of more complex structure, tend to change their behavior according to the very thick of the analyzed layer.
However, most of the following heat transfer principles linearly. Therefore, I advise to measure small and large-scale thermal changes, applying the heat transfer law, the correction factor.
For very thin layers, the best analysis can be by the use of electronic microsensors, like high sensitivity thermistors. Or, infrared emissions from the surface analyzes. However, it is necessary in this case, analyzing the thermoradiation emissivity of the material.
The problem you had specified can be solved as a 1D steady state heat conduction in composite walls. Initially you can do so to determine the approximate values of intermediate temperatures. To make it much more perfect, you must add contact resistances between liners and wall. You must also have heat transfer coefficient of gas contained inside and outside the furnace.
In general the formulation is Q is ratio of overall temperature change to the overall resistance.
Overall Temperature change - Gas temperature inside furnace - Ambient temperature outside furnace
Overall resistances = Resistance of liners + Resistance of refractory wall + Resistance due to convection inside furnace + Resistance due to convection outside furnace + Contact resistances.
Assume that Q is same throughout all materials and also through gases.
Hope you had got the basic calculation.
The heat transfer by conduction, materials with isotropic behavior, on a cobblestone object, follows the following law: q'/A = k.ΔT/x
where:
q'/A --> is the amount of heat per second per unit area, transferred by wall square meter (SI); W/m2;
∆T --> variation of temperature, ˚C;
x --> wall thickness, m.
For this finality, we assume that the wall is a parallelepiped.
Another principle we take as a viable is the fact that it presupposes that the thermal conductivity, k, of the material does not depend significantly on the temperature. In this case, it behaves linearly.
But if the case for further and detailed study should take into account that: k (x) = f (x) = ko (1 + Cx).
Yes. Mr. Elvelto is right. And k(x) is indeed a function of temperature in Kelvin and formula he had listed is for solid and there is another formula for k for fluids if it varies with respect to temperature.
Indeed, the Cx = f(T), where T is the used in Kelvin.
Therefore, ΔT in ºC is equal in the Kelvin scale.
For the liquid heat transfer, is common consider only the convection phenomena, but, in most deep searching we can put the liquid conduction in the seek.
Remember that convection only occurs in the presence of gravity, or another ponderability force field.
What if there are three layers and the heat transfer is temperature dependant and differing for all materials? What about the geometry? If it is a cylinder can we assume that a 90 degree segment is the same as all other segments?
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