Dear Hakeem,

Unfortunately, the critical thickness of an insulation exist only in spherical and cylindrical heat barrier - the thermal resistance reaches a minimal value for that walls. When you assuming that constant heat conductivity through the plane wall, with respect to the temperature, you can choose the thickness of the insulation whatever you want - more is better. You must to choose maybe different parameter to optimize the insulation thickness. I think economical aspects will be enough.

Best regards,

PS

Thanks Przemysław for the input - For optimizing the insulation thickness, which parameter/process/approach is better...because glass wool is a bit costlier insulation material and its optimization is must when applied in industrial scales

Finally I have got it - Using the 1D plane wall assumption in all 6 sides and getting the heat lost by giving some thickness of insulation.

I agree with the explanation given by Przemysław. If there is no constraint for thickness then only the cost minimization would be the objective for optimization.

I would say that if the side of the cube is not large (say, one order of magnitude) when compared to the expected insulation thickness, then the heat lost through the vertices may be not entirely negligible. It doesn't seem to be your situation. But if it was, then you may want to consider the insulation layer around each of the 8 vertices of the cube as a problem corresponding to 1/8 of a sphere. Then, if you forget about the six plane walls (which is the part of the problem that is well understood), you can look at the problem as that of a sphere! Determining the critical radius for this, could give you some reference value to work with for your cube.

I agree with Daniel Vaz. Interesting question. I think there may also be a critical thickness associated with the cube, IF you keep in mind that the edges and corners are where most the the heat wil be lost to conduction.

Assume you are using flat insulation. To control loss from the edges and corners of this meter cube, the flat sheets must be larger than 1m by 1m, and their edges cut at 45 degrees for a snug fit. This creates a larger surface area for the insulated cube than the 6 sq m surface area of the cube itself. Hence, the potential for more convection than conduction loss.

Or put another way, if one does not overlap the flat insulation at the edges and the corners, then simply adding 1 sq m sheets of thicker and thicker insulation will show rapidly diminishing returns. Much more rapidly diminishing returns than if optimal thickness were only a matter of added cost of insulation.

(Notice I didn't yet derive the critical thickness equation!)

If the structure is an almost a cube, it could bring near to a spherical cap. And it is possible to look for the critical radius for insulation.

Critical thickness of insulation exists only when the surface area increases with the increasing thickness like in cylidrical and spherical shapes. For plane walls the surface area is invariant with the insulation thickness hence there is no critical thickness

Thanks everybody for the inputs

Special Thanks to Fellah for the conceptual explanation

In your case Fellah's answer is not accurate. If you just increase insulation thickness in the normal direction from the cube walls, you will add the following surface area on a given side of the cube: thickness*4 m²

(this calculation is for a 1*1*1 m³ cube)

Personnally, I would go with Daniel Vaz's answer and work it out as a sphere problem, considering you put insulation at the corners, which seems much more logical than only in the normal direction.

One way of treating the planar wall case is to consider very large or infinite radius for a cylindrical wall and derive the equation but this method has mathematical difficulties to deal with infinite quantities ( which has to expressed as reciprocals to make them zero).

Critical thickness of insulation means maximum heat loss with min thermal resistance. This concept is useful in the situation when surface area of the base object increases with the insulation, i.e., cylindrical and spherical systems.

In case of plane wall or cuboid, the addition of insulation layer over the base surface will not make any effect on surface area and hence reduce the heat loss from the system irrespective of the thickness of the insulating material.

Moreover, in industry, critical thickness of insulation has little role since the dimension of the base surface is in macro-size which will definitely reduce the heat loss with insulating layer. Optimum thickness of insulation will play a significant role which involve cost of the insulation, labor and counting the period of its last.

Hemantkumar Mehta, do you agree with me that when you add insulation on the surfaces of a cuboid, the result is a bigger cuboid? => increased heat loss area.

My last contribution is valid only for one dimensional plane wall problem. For cubic shapes I agree with Mr. Gregoire

Dear Prof. François Grégoire,

Yes I do agree with your answer and I will go with the statement of Prof. Giuma Fellah about my contribution.

Critical thickness concept arises due to increase in surface area as we go on adding insulation like in cylindrical and spherical systems, there by reducing convective resistance along with increase in conduction resistance.But for a plane wall adding insulation only increases conduction resistance without changing convective resistance .Hence this is not applicable.

However in case of Cube when we add insulation as side increases the total surface area increases hence the concept of critical thickness comes in to picture.

we cannot use the insulation on faces treating them as 6 isolated walls.

Thanks to all for all the inputs provided

Its really a good discussion on the concept of critical thickness for cube/cuboid

Apart from the cost of the insulation, there is no critical thickness of insulation in flat plane wall. But after a certain thickness the redution in heat transfer by ussing more insulation is insignificant.

Yes Kharseh you are correct - the "certain thickness" mentioned by you needs a proper term other than critical thickness...

I got to see this thread just now and felt the need to add a clarification. Hakeem considers a 1m x 1m x 1m cubical heat exchanger and wonders about whether a critical thickness consideration exists for the edges (similar to a cylindrical segment) and corners (like a spherical one).

Hemant kumar Mehta has said all that was necessary in these words: "IN INDUSTRY, CRITICAL THICKNESS OF INSULATION HAS LITTLE ROLE SINCE THE DIMENSION OF THE BASE SURFACE IS IN MACRO-SIZE WHICH WILL DEFINITELY REDUCE THE HEAT LOSS WITH INSULATING LAYER. OPTIMUM THICKNESS OF INSULATION WILL PLAY A SIGNIFICANT ROLE WHICH INVOLVE COST OF THE INSULATION...."

For the benefit of the young researchers who must surely be following the posts here, I wish to offer additional explanation. Just take a look into any introductory book on Heat Transfer (say, Cengel or Arpaci) and you will find that the critical radius concept applies only to VERY small sizes, typically a few millimeters. To reach a critical radius in the range of meters, you will require impractically large values of thermal conductivity (no more an insulation!) or impossibly low values of the external heat transfer coefficient. So Mr. Hakeem need not worry about critical insulation thickness unless his heat exchanger is, say, a 5 mm x 5 mm x 5 mm cube!

I agree with Vijay's explanation, even for cylinder and sphere, critical thickness is only for small diameter tube or sphere r=k/h. For large system (even order of an inch) adding insulation decreases the rate of heat losses.

Really it is an eye opening discussion. Normally pepople will think this concept exits for only for sphre and cylinders

I agree with Vijay Raghavan and Abdulmajeed Mohamed edxplanations, because such explanations we could find in many books of Heat Transfer, say Cengel, Kakac..., where the critical raduis concept applies only to very small sizes!

So many things have been discussed but for outer surface temperature is there any guideline based on outer surface temp of insulated body?

Sorry Mr. Kanaujia, your question is not clear, except to perhaps to yourself. Could you take another look at the wording of the question and make it intelligible to other readers?

There is a difference between optimum thickness and critical radius of insulation.

Critical thickness of insulation is the critical point of thermal resistance curve along with radius of sphere or cylinder shapes surface and not exist for plane surface. In plane surfaces as thickness of insulation increases the conduction resistance increases and no effect on convection resistance.

Optimum thickness of insulation is based on minimizing the total cost, running cost and fixed cost. Running cost of heating or cooling . Fixed cost of thermal insulation thickness.

It's such a good discussion about the critical insulation. I have a similar query for my refrigerator whose current insulation thickness is not sufficient. I want to know what should be the optimum thickness of that insulation to prevent the heat loss from the cabinet , so that I can retain the cold temperature for longer durations. The refrigerator shape is of cuboid of aprrox 1 cu.m volume.

2k/h-a/2 is the answer. a is the side length of the cube. If a goes to 0 the critical radius of the sphere is recovered.

Yes, when surface area enhances with increase of insulation, net heat transfer by convection increasue since surface area exposed to surrounding fluid enhances, irrespective of geometry.

For a tube having cubical cross section (a*a), and length L, initial surface area, without insulation (Neglecting the area of inlet and outlet is) is 4*a*L. When t thick of insulation is added, surface area will become4*(a plus t)*L.

However, if we consider a plane surface, and heat is only transferred transverse to the plate; then surface area(through which) conduction happens remains constant, a*L. That's why, the notion is planar geometry does not have critical thickness issue. But people must understand the assumptions as I had briefly elucidated above.

So, ofcourse, with increase of thickness, with increase of surface area, conduction resistance will increase and convective resistance must decrease.

If temperature gradient from surface remains equal in all directions, surface temperature of tube, surrounding temperature as well as the heat transfer coefficient of fluid remains constant with thickness of insulation remsins same throughout; a simple calculations will show that,

Critical thickness equals (k/h-a)/2.

Thank you.

Dear Sanko. Of course there are semi-spherical and semi-cylinderical structures at edges and vertices.

Dear Hakeem Niyas

For flat plates there is not a critical thickness of insulation.

As in the case of a circle, the critical radius of insulation is (k/h).likewise the shortest length in the case of a square will be 0.7994(k/h)