Effect of thickness on heat transfer/insulation

[Canuck156]

Hi,

I'm currently doing an experiment to investigate the effect that changing the thickness of insulation has on cooling curves. However, I do need some theory to compare the results to. Does anyone know anywhere that I could find such information, or actually know themselves what effect changing the thickness should have?

Thanks.

 

[Doc Al]

heat conduction

Perhaps this will get you started. In general, the rate of heat transfer by conduction is inversely proportional to the thickness of the material:
$$\frac{\Delta Q}{\Delta t} = \frac{k A \Delta T}{d}$$
where ΔQ/Δt is the rate of heat flow, ΔT is the temperature difference, k is the thermal conductivity of the material, A is the area, and d is the thickness.

(Do a web search on heat conduction to find plenty more information.)

 

[Canuck156]

Thanks.

Does this mean that if all variables are kept constant except for the thickness of the insulation, and the temperature inside the insulated area is modeled by $$T_{n}=T_{O}\times{e}^-^k^n$$ (The insulated area is heated up, and then the air is allowed to cool) that k will by directly proportional to the thickness?

 

[Gonzolo]

How would you simoustaneously maintain both ΔQ/Δt and ΔT constant?

 

[russ_watters]

How would you simoustaneously maintain both ΔQ/Δt and ΔT constant?

In the summer, with air conditioning.

edit: To make that sound a little less snide, let me explain. Obviously, if you don't have air conditioning, the air in your house will slowly increase its temperature to match the outside temperature: ΔT decrease to zero and ΔQ/Δt will follow. On the most basic level, the purpose of an air conditioner is to maintain a ΔT between inside and outside. Constant ΔT and ΔQ/Δt requires an another term: another ΔQ/Δt. Energy enters your house through the wall and leaves your house through the air conditioner.

 

[Canuck156]

Sorry, I think that what I said was not exactly what I meant. By keeping 'all other variables' constant, I meant that I would keep k and A constant, while varying d.

Sorry for the confusion.

 

[Canuck156]

Another question regarding the modelling of this situation:

If two different materials were used, (ie. 1 layer of material A, and 1 layer of material B, pressed together) as the barrier, how would the equation $$\frac{\Delta Q}{\Delta t} = \frac{k A \Delta T}{d}$$ need to be modified to compensate for that?

 

[Doc Al]

composite layers

For two slabs of material:
$$\frac{\Delta Q}{\Delta t} = \frac{A \Delta T}{d_A/k_A + d_B/k_B}$$

 

[Canuck156]

Thanks, that helps a lot. Is there an internet site or book that contains information on dual layer conduction? I've been looking but I haven't been able to find one.

 

[TurbulentCFD]

Thanks, that helps a lot. Is there an internet site or book that contains information on dual layer conduction? I've been looking but I haven't been able to find one.

Reference the Fundamentals of Heat and Mass Transfer by Incropera and Dewitt.

 

[REINHARD]

It gets confusing sometimes...When composite slabs are involved.If 2 or 3 slabs are involved then it won't be a problem.But I had encountered some really tough problems on this.So I think the best thing to do is to find somekid of anology btw Electric circuts and the Slab-Systems...

So by putting R=d/kA,where R is Thermal resistance.
Then everything is like that of Eletric circuits...Ohms law holds good for thermal conduction also.

 

[Canuck156]

Ok, I understand how to do that now, but I basically now have two formulas. From my experimentation I have:
$$T_{t}=T_{Difference}\times{e}^-^k^t$$, where $$TT_{Difference}$$ is the initial difference in temperature, and $$T_{t}$$ is the difference after t seconds.
and from the theory I have:
$$\frac{\Delta Q}{\Delta t} = \frac{A \Delta T}{d_A/k_A + d_B/k_B}$$

I am trying to find a mathematical relationship between the value of k in formula one, and the value of DQ/Dt in the second equation. Is it possible to do this using something like:
dQ/dt=dQ/dT*DT/dt, or am i on the wrong track?

Thanks.