Insulation thickness/surface temperature - Heat Transfer & Thermodynamics engineering (2024)

Further to your 1st question, both you and "some people" are saying the same thing.

The heat transfer through insulation (and just through insulation) is governed by conduction.

Q = 2*pi*L*k*DeltaT/ln(d2/d1)

Where:

Q = heat flow through insulation [W]
k = insulation thermal conductivity [W/(m°C)]
DeltaT = temperature difference between outer and inner surface of the insulation [°C]
d2 = outer insulation diameter [m]. d2 is equal to the pipe diameter + twice the insulation thickness
d1 = inner insulation diameter [m]. d1 is equal to the outer pipe diameter

This is an increasing function (just differentiate with respect to d1) and so heat increases as pipe diameter increases.

Further to your 2nd if you could post your spreadsheet maybe someone could help you.

Dear Ione,

thank you very much for your answer.
Thas correct, it follows my idea and my analysis of the equation!

In relation to the second point, i found the problem. I was using a ABS() function to give positiva values, and it was messsing with the results. I removed it and the results now are fine!!

thanks, i think it's all cleared!!

Marcosdias,

As you know the overall heat transfer coefficient involves also convection at the inside pipe surface, conduction trough the pipe thickness, conduction through insulation thickness and convection at the outside insulation surface (pls. see the attached pdf). The heat transfer is the results of four processes (thermal resistences in series), and also inside and outside convection coefficients (hi and ho) are influenced by the pipe diameter.

• http://files.engineering.com/getfile.aspx?folder=01992e26-0511-407e-8b00-4d

Then, of course, there is another strange result. If the ratio k/h of conductivity k (W/m.K) to the external convection coefficient h (W/m².K), is greater than the outer radii (including insulation) an increase in insulation thickness will cause an increase in heat transfer. Generally applicable to wires and very small diameter tubes.

Yes, there is a particular radius of the insulation(critical radius = k/h) at which heat transfer reaches its maximum. For this value the slope of the heat flow is equal to zero.

• http://files.engineering.com/getfile.aspx?folder=af389be4-d8b5-4e85-b8ec-86

As a side -

There is a good insulation program available for free to calculate all these things:

http://www.pipeinsulation.org/

Dear AllHandlesTaken,

I know, but the 3E Plus is a bit strange it gives strange results i have gaven up of using it.

marcosdias,

Try the link below

http://www.kaimann.de/en/service/kaicalc.html

There is a freeware program for insulation which works well (it deserves a star!)

Dear ione,

Thank you for the link, but i can't download it because Kaimann links are broken

I will try again later

marcosdias,

The link has been out of service for a couple of days, but now it is operative: give it a shot!

Great IONE,

thanks for the post!
I'm downloading it.
I will give my opinion later.

P.S.: My excel worksheet is finished and it's working fine.
Maybe i will share it later, when i have time to translate it to english, becaus is in Portuguese...

Thanks!

") What some people say is that as lower is the pipe diameter the less would be the insulation thickness, to reach the same surface temperature. Well, what i found is the opposite, as i insert bigger diameters, the insulation thickness increases very very much, even to stupid values...
We can see that also from the equations: as we increase the pipe diameter for the same insulation thickness, the LN(r2/r1) decreases and the heat transfer coeffecient increasesm, therefore the heat loss also increases.
I would like to see some comments."

"Some people" are wrong and you are right
Another way of looking at this is the thermal resistance through the insulation is
ln(ro/ri)/(2piK)
Now keeping the temperature constant at the outside surface demands that the resistance be the same. Looking at the above expression it is seen thatthe resistance is constant when ro/ri ic constant. So for increasing ri the difference ro-ri must increase, proving your results.

Marcosdios,

Go ahead and post it as is for those of us who speak Portuguese.

rmw

dear davefitz,

I'm quite confused about what you've said.

In the radiation heat transfer equation, if you increase the emissivity, you increase the heat transfered. so, how do you lower the insulation thickness??

Thanks for the post!

I think what davefitz suggested (but maybe I am wrong and davefitz could better explain) it is related to min thickness to prevent condensation.

Take a look to the link below.

http://books.google.it/books?id=IMY8nBoaYZQC&pg=PA106&lpg=PA106&dq=jacket+emissivity+insulation&source=bl&ots=Jglm6R6axW&sig=V95O6vxMjPU3g-JLKzjTYsjfrBA&hl=it&ei=5VJYS7KvGc7dsgbh54TZCw&sa=X&oi=book_result&ct=result&resnum=9&ved=0CDMQ6AEwCA#v=onepage&q=jacket%20emissivity%20insulation&f=false