Heat loss from non-lagged pipe -basic calculation 3

[mrtangent]

Does anyone have a simple rule of thumb that they could suggest for an estimation of the pipe heat loss for a un-lagged pipe.

If we send a product down a 1200 meter line to a storage tank at 40°C, I think it will be cooled down to atmospheric temperature based on a calc I've done but I would appreciate if anyone could suggest some resources to check my numbers. I adapted the Kern method - but there must be some rule of thumb HTC I could check but I've not found them in perry etc.

 

[TD2K]

Kern (page 215) for horizontal pipes gives the external film coefficent of 0.50 * (dT/do)^0.25 for the convective heat loss coefficient (US customary units, dT deg F, do inches, hc BTU/hrft2F). Don't forget to include the radation heat loss.

Wind can increase the convective heat losses by several times.

 

[CRG]

Hello James,

You can verify your calcuations with a free program from the North American Insulation Manufacturers Association (NAIMA).
Download program from:

http://www.pipeinsulation.org/pages/home.html

Regards

 

[Montemayor]

CRG:

Thanks for the very helpful post. This program easily voids the QuickBasic program I wrote for myself 7 years ago for the same purpose as expressed by James. It's nice that the insulation suppliers have now gotten on the same page with us engineers who have to calculate and specify the product to buy.

TD2K is right on the money when he cautions about the atmospheric wind effect on convection heat losses - in most cases, it dominates the heat transfer results.

Art Montemayor
Spring, TX

 

[25362]

Although all the comments brought by my predecessors are valuable, it seems that james1030bruce is speaking of an unsteady state heat transfer situation.

If it is so, then the problem is more complex than it seems, and graphs, as the Gurney-Lurie diagram for heating or cooling of cylinders, would be applicable to determine the time needed to reach a certain temperature at the end of a certain pipe length when there is turbulent fluid flow, as well as at a certain distance from the pipe center when stagnant.

I'm not sure, but probably specialized heat transfer books may bring simplified solutions as requested by James.

 

[CRG]

Does anyone have a simple rule of thumb that they could suggest for an estimation of the pipe heat loss for an un-lagged pipe?

Look at the 1200 meter pipeline 50 meters at a time. Use the 3E Plus program to evaluate each segment assuming constant temp. Use the heat loss information to calculate the average temp drop for that segment. Use the outlet temp for the next segment. Do this 24 times and you will have an estimation of final temperature at the outlet of the line. Because the computer program does the steady state heat loss calculation, it should be fairly quick to analyze all of the pipeline segments. Seems to me this should be a fairly quick way to get an accurate estimation. Also, as the solution is being iterated, it can be seen if your segment length need to be increased or decreased. For most heat loss calculations of this type, inside film resistance can be ignored because they are so small compared to the rest of material. For bare steel pipe, it is the outside air film that dominates the solution.

 

[25362]

The section-by-section analysis, as CRG says, is a practical way to make heat loss, or gain, estimates, especially on flowing lines as long as this one, along which the external atmospheric (weather) conditions may change.

It is true that for most heat loss or gain calculations the internal resistances can be neglected, but not for all circumstances as one may imagine. For example, when pumping oils.
In these cases the convection HT coefficients may have the same order of magnitude, for example, 0.1-0.2 kW/(m².K).

For the case of a stagnant fluid to estimate the time it takes for a given internal temperature to reach a given distance from the center of (an infinitely long) pipe, better use diagrams as I mentioned above.

 

[mrtangent]

Thanks for the help.

CRG - The insulation program gives me the accuracy i need for the purposes of my calculation.

 

[magicme]

hi there

much thanks to CRG for that link.

as an aside......

the exit temperature of the internal fluid would be calculated by the following equation, which uses the log mean delta T method....

Texit = Tambient -(Tambient - Tinlet) * exp ( -U * A / mdot / Cp)

where
U is the overall heat transfer coefficient of the pipe wall..

U = 1 / (1 / hinternal + 1 / hexternal + t / k)
t= wall thickness
k = conductivity
h = convection coefficients

A = surface area of the inside wall of the pipe
A = pi * Dinternal * Length

mdot = mass flow rate inside the pipe
Cp = internal fluid specific heat

"exp (x) " means e^x

watch the units of mdot and Cp.....may need factor of 1000 in that multiplication.

daveleo