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Insulation thickness/surface temperature

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mfqd13

Mechanical
Sep 27, 2007
99
Hi,

I have a Excel sheet wich calculates the insulation thickness, so that the surface temperature reaches, a user defined temperature, for example 40ºC, for safety reasons.
It happens 2 things that i consider quite strange:

1) 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.

2) i have a little macro wich uses the goal seek function, and it change increases the insulation thickness until the surface temperature lowers to the specified value: 40ºC. The problem is that excel finds a value wich represent a surface temperature value, not exact. The deviation error goes from 5 to 20%. This is very strange and is the 1st time that it happens to me. Manually, i have no problem to find the exact value. What maybe occuring?

Thanks
 
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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-8b00-4d84f2915daa&file=Overall_heat_transfer_coefficient.pdf

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.
 
Dear AllHandlesTaken,

I know, but the 3E Plus is a bit strange it gives strange results i have gaven up of using it.
 
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 that the 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
 
another trick that can be used to minimize insulation thickness: standard aluminum pipe lagging may have an emissivity of 0.10, but if you use anodized aluminum lagging, the emissivity is 0.90. The use of anodized lagging would allow much thinner insulation based on radiative cooling.
 
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!
 
The increase in heat transfer rate at the surface serves to lower the surface metal temperature. In the case where the objective is to prevent harming a worker due to a high surface temp, using anodized aluminum will reduce the required insulation thickness. However, the loss of heat from the piping to the environment will increase, causing the root process to consume more energy.
 
It is true that low lagging emissivity increases outer surface temperature at a given insulation thickness. Consequently increasing the jacket emissivity the outer surface temperature drops. Anyway the trade-off is quite expensive as the heat dissipated increases. I would prefer to use warning signals to inform personnel about risks related to hot surface contact.
 
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