Critical insulation thickness

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Saad85

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Hi all,

I am confused with critical insulation thickness in MERM page 32-11 , the question how the insulation increase and decrease heat loss?

 
Hi all,

I am confused with critical insulation thickness in MERM page 32-11 , the question how the insulation increase and decrease heat loss?
The addition of insulation makes it thicker, so the resistance due to conduction increases. But, at the same time, as you add insulation the outer surface area of the insulation increases, so the resistance due to convection decreases. So as you add insulation from zero thickness you have two competing effects: increasing conduction resistance versus decreasing convective resistance. 

Imagine you have a bare pipe and add a very thin layer of insulation. For very thin layers, there is actually an increase of heat transfer because the heat transfer increasing effect of the bigger surface area is greater than the heat transfer reducing (insulating) effect due to the insulation layer thickness. If you add another thin layer you make it worse, and another thin layer also makes it worse until at a certain thickness the insulating effect of the added thickness becomes dominant. After this point, adding layers contributes more to the increase the conductive resistance than it does to decrease the convective resistance. That point where this crossover occurs is the "critical insulation thickness"

 
Thanks for the feedback let assume we have Propane storage tank and there is insulation around the tank if we add 120% more insulation what happened to the heat transfer decrease or increase?

 
It will depend on the insulation conductivity and the convective heat transfer coefficient. In your case we know for sure that the resistance due to conduction will increase (due to the increase in thickness) and the resistance due to convection will decrease (due to the increase in surface area), but we can't tell for certain which change in resistance will be more. If the increase in conduction resistance is more than the decrease in convection resistance, the rate of heat transfer will decrease. If the increase in conduction resistance is less than the decrease in the convection resistance, the rate of heat transfer will increase. 

 

 
It will depend on the insulation conductivity and the convective heat transfer coefficient. In your case we know for sure that the resistance due to conduction will increase (due to the increase in thickness) and the resistance due to convection will decrease (due to the increase in surface area), but we can't tell for certain which change in resistance will be more. If the increase in conduction resistance is more than the decrease in convection resistance, the rate of heat transfer will decrease. If the increase in conduction resistance is less than the decrease in the convection resistance, the rate of heat transfer will increase. 

 
Exactly. 

Its impossible to tell what will happen unless you specify:

1. Current thickness of insulating layer.

2. Thermal conductivity of insulating layer.

3. Convective coefficient. 

 
True.

But assuming normal circumstances (normal values for insulation and convective film coefficient), you can almost bet that the heat transfer will decrease in this case. The critical radius for this application is most likely very small, so the new insulation will most likely increase the conduction resistance more than it decreases the convection resistance and thus the rate of heat transfer decreases.

 
Let us assume the thermal conduction of insulation  is 0.016, convective coefficient is 3 thinkness of insulation layer is 2 in , the diameter of tanks is 4 in 

 
Well, I wouldn't call a 4in diameter sphere a "tank" :eek:

I also notice you are not writing units next to some numbers. Not keeping track of units is probably the biggest hindrance to success in the PE exam. Please start doing this consistently.

So I'm ASSUMING these numbers have consistent units and we can calculate the sphere's critical radius as r = 2k/h = 0.0106 (IM GOING TO ASSUME THIS IS INCHES) Never work like this either studying for, or taking the test. We have no idea what units this "radius" is in.

If the insulation is currently 2 inches thick on a sphere of radius 2 inches, means the current radius is 4 inches. This is way larger than the critical radius; therefore any addition of insulation will decrease the heat transfer rate.

 
Well, I wouldn't call a 4in diameter sphere a "tank" :eek:

I also notice you are not writing units next to some numbers. Not keeping track of units is probably the biggest hindrance to success in the PE exam. Please start doing this consistently.

So I'm ASSUMING these numbers have consistent units and we can calculate the sphere's critical radius as r = 2k/h = 0.0106 (IM GOING TO ASSUME THIS IS INCHES) Never work like this either studying for, or taking the test. We have no idea what units this "radius" is in.

If the insulation is currently 2 inches thick on a sphere of radius 2 inches, means the current radius is 4 inches. This is way larger than the critical radius; therefore any addition of insulation will decrease the heat transfer rate.
Those values are probably in Btu/hr-ft-F and Btu?hr-ft^2-F respectively, so the critical radius will be in feet. But your final conclusion stays absolutely correct.

Saad, Just take a look at the critical radius formulas for both cylinders (r = k/h) and spheres (r = 2k/h) and you'll notice that with typical values of insulation conductivity (<0.1 Btu/hr-ft-F) the critical radius is always very small, so any added insulation will almost always decrease the rate of heat transfer.

 
But again, original radius, insulation thickness, conductivity, and convective coefficient are a must to be able to tell for sure.

 
Thanks for the feedback let assume we have Propane storage tank and there is insulation around the tank if we add 120% more insulation what happened to the heat transfer decrease or increase?
When do you plan to take the PE exam?

 

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