Posted by *daestrom* on January 17, 2006, 1:45 am

*> How do you figure out the thermal loss of the pipe?*

*> My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.*

*> So, 25/R * Temp_Diff = loss in BTU/HR*

*> For R10 at 50 degree that's only 125 BTU/Hr. Calculating R value for *

*> circular vs flat perplexes me... My guess is that the R value would be *

*> lower than what you normally expect per inch.*

R values are calculated from the thickness of insulation and the thermal

conductivity for flat, 'infinitely large' panels. That is to say, the

calculation assumes the heat is 'flowing' in a straight line, straight

through the material.

R = thickness / (conductivity * area)

Sometimes, for things like insulating panels, they omit the 'thickness' as

it is implied in the particular panel. For example, some rigid foam has a

conductivity value of 0.2 (BTU-in / hr-F-ft^2). So a two inch piece, would

have an R value of 2/(0.2*1) = 10 F-hr/BTU-ft^2. With a temperature

difference of 30 F, and an area of 5 ft^2, the heat flow would be (30 F * 5

ft^2) / (10 (F-hr-ft^2/ BTU)) = 15 BTU/hr

For pipes, there is a formula *almost* as simple as for flat surfaces.

R = ln(r2 / r1) / (2*pi * conductivity * length)

Where r2 is the outer radius of the insulation, and r1 the inside radius.

But you have to take care to measure the length and conductivity all in the

same units. The same foam (conductivity of 0.2 (BTU-in/hr-F-ft^2)) has a

conductivity of 0.2/12= 0.0166 (BTU-ft/hr-F-ft^2). If a 3/4 inch layer is

wrapped around a 1 inch pipe that is 30 feet long, then the inner radius r1

would be 0.5 in, and the outer radius would be 1.25 in.

R = ln(1.25 / 0.5) / (6.28*0.0166*30) = 0.292836 F-hr/BTU

Sometimes the length is omitted to get an R-value linear foot. Similar to

getting an R-value per square foot for flat surfaces. The same example

would be 8.785 'R' - ft.

If the OP's piping is insulated like this, and is 200 ft long (100 ft each

way), then the total R value would be 8.785 / 200 = 0.043925 F-hr/BTU. If

the piping is a modest 30 F warmer than the soil immediately surrounding the

pipe, the losses would be...

Q = 30F / 0.043925 F-hr/BTU = 683 BTU/hr

Hope this helps

daestrom

Posted by *daestrom* on January 17, 2006, 2:03 am

*> How do you figure out the thermal loss of the pipe?*

*> My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.*

area = 100 ft * pi * (3/4)/12 = 19.6 ft^2

*> So, 25/R * Temp_Diff = loss in BTU/HR*

*> For R10 at 50 degree that's only 125 BTU/Hr. Calculating R value for *

*> circular vs flat perplexes me... My guess is that the R value would be *

*> lower than what you normally expect per inch.*

If 2 inch thick closed-cell foam gives you R10 for a flat panel (thermal

conductivity 0.01666 BTU/hr-F-ft), but instead you wrap it around the pipe,

the trick is figuring out whether to use the surface area of the 3/4 inch

pipe (19.6 ft^2), or the surface area of the outside of that foam wrapper

(area = 100ft* pi * (2.75)/12 = 72 ft^2). Or somewhere in between.

The derivation is complicated, but the results are....

'R' = ln(2.375/0.375) / (2*pi*0.01666) = 1.76334

Q = 100 ft * 50F / 1.76334 = 283.6 BTU/hr

daestrom

Posted by *Jeff Thies* on January 17, 2006, 3:23 am

daestrom wrote:

*> *

*>>How do you figure out the thermal loss of the pipe?*

*>>*

*>>My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.*

*>>*

*> *

*> *

*> area = 100 ft * pi * (3/4)/12 = 19.6 ft^2*

Actually, I was thinking that the outside diameter of 3/4" pipe was more

like 1". No matter...

Thanks for this.

It's interesting to note that for relatively thin 1" insulation (like

the pipe wrap at stores). Smaller pipes insulate better. A 1" OD pipe

with 1" insulation has 82% of the R value that a .75" OD pipe with the

same 1". This gives me thought when I redo my plumbing (also the lesser

volume of water in the pipe cooling)!

Cheers,

Jeff

*> *

*> *

*>>So, 25/R * Temp_Diff = loss in BTU/HR*

*>>*

*>>For R10 at 50 degree that's only 125 BTU/Hr. Calculating R value for *

*>>circular vs flat perplexes me... My guess is that the R value would be *

*>>lower than what you normally expect per inch.*

*> *

*> *

*> If 2 inch thick closed-cell foam gives you R10 for a flat panel (thermal *

*> conductivity 0.01666 BTU/hr-F-ft), but instead you wrap it around the pipe, *

*> the trick is figuring out whether to use the surface area of the 3/4 inch *

*> pipe (19.6 ft^2), or the surface area of the outside of that foam wrapper *

*> (area = 100ft* pi * (2.75)/12 = 72 ft^2). Or somewhere in between.*

*> *

*> The derivation is complicated, but the results are....*

*> *

*> 'R' = ln(2.375/0.375) / (2*pi*0.01666) = 1.76334*

*> *

*> Q = 100 ft * 50F / 1.76334 = 283.6 BTU/hr*

*> *

*> daestrom*

*> *

*> *

Posted by *daestrom* on January 17, 2006, 10:13 pm

*> daestrom wrote:*

*>>*

*>>>How do you figure out the thermal loss of the pipe?*

*>>>*

*>>>My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.*

*>>>*

*>>*

*>>*

*>> area = 100 ft * pi * (3/4)/12 = 19.6 ft^2*

*> Actually, I was thinking that the outside diameter of 3/4" pipe was more *

*> like 1". No matter...*

Guess it all depends on the pipe wall thickness. And that of course depends

on the material.

*> Thanks for this.*

Your welcome

*> It's interesting to note that for relatively thin 1" insulation (like *

*> the pipe wrap at stores). Smaller pipes insulate better. A 1" OD pipe with *

*> 1" insulation has 82% of the R value that a .75" OD pipe with the same 1". *

*> This gives me thought when I redo my plumbing (also the lesser volume of *

*> water in the pipe cooling)!*

Yes, it can be pretty 'contrary'. If you look at really small diameter

pipes or heated wires, you run into a situation called 'critical insulation

thickness'. For really small diameters, the insulation increases the

surface area more than the insulation reduces heat flow, and adding

insulation actually can *increase* the heat losses. But that's not a

problem with sizes in this range.

Of course, smaller diameter piping means more pressure losses and/or less

flow. As with almost all things, it's a compromise.

daestrom

Posted by *Gary* on January 17, 2006, 11:32 pm

Hi Jeff,

I don't have my notes handy, but I used the same equation that Daestrom mentions

for heat loss through cylinderical insulation. I ignored the pipe wall, and

assumed the insulation was a cylinder, and that there was a constant temperature

around the outside of the insulation.

q = 2 Pi k L (Ti - To)/ ln(Ro/Ri)

Where i is inner and o is outer

None of this is true, since the insulation was rectangular, and there are two

pipes embedded in the same block of insulation, and there is a temperature

gradient as you go down, and the ground may warm up a bit around the pipes over

time? But, it seemed like it would still give some guideance for how much

insulation is needed. Most of the assumptions seem conservative.

The things I wanted to confirm were:

1 - The heat loss while hot water is being pumped is a small percentage of the

heat that the pipe is transfering

2 - That when the pipe sits idle for a long time (hours) and loses all of its

heat that I could afford the total loss of the heat that was in the water in the

pipe -- ie that there is not a lot of heat stored in the pipe itself.

I think that both are OK -- the first argues for thick insulation and a small

pipe OD to minimize heat loss, and the 2nd calls for small pipe diameter to

minimize the amount of heat stored in the pipe itself. So, I set the pipe

diameter to the minimum I thought could handle the heat flow I needed from the

storage tank to the house without letting the pumping losses get too high. The

3/4 inch diameter and the block insulation (I believe) statisfies the two

conditions.

I did just notice that my Heat Transfer book does have a pretty simple way to

handle various shapes buried in an isothermal media -- I'll have a go at this

and see if it changes much.

Gary

Jeff Thies wrote:

*> How do you figure out the thermal loss of the pipe?*

*> *

*> My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.*

*> *

*> So, 25/R * Temp_Diff = loss in BTU/HR*

*> *

*> For R10 at 50 degree that's only 125 BTU/Hr. Calculating R value for *

*> circular vs flat perplexes me... My guess is that the R value would be *

*> lower than what you normally expect per inch.*

*> *

*> Cheers,*

*> Jeff*

*> *

*> *

*> *

*>> wp wrote:*

*>>*

*>>> I live on the coast of Maine where the winters can get pretty harsh*

*>>> and the frost, in a bad year, will get down 3-4 feet in the soil. I'm*

*>>> thinking about putting up some solar panels for hot water heating and*

*>>> the best spot for sun is about 100 feet from my house. I have access*

*>>> to a backhoe so digging the trench deep enough is no problem, but I'm*

*>>> wondering about a couple of things. First, will I lose too much heat*

*>>> as the heated water travels that distance underground? Second,*

*>>> assuming that it is feasible, does anyone have any experience with*

*>>> this type of application, ie. what sort of pipe to use, how to*

*>>> properly insulate them and how to keep any movement underground from*

*>>> bending the pipe to breaking point. My plan is to put the panels on a*

*>>> framework that can manually pivot with the sun (I'm home most of the*

*>>> day so I should be able to get good facing most of the time) and I*

*>>> don't mind oversizing the system to compensate for losses underground.*

*>>> Any suggestions would be welcome.*

*>>> Thanks!*

*>>> WP*

*>>*

*>>*

*>>*

*>> Hi,*

*>>*

*>> You might want to take a look at how I am doing my buried pipes -- *

*>> sounds like*

*>> my situation is pretty much like yours.*

*>> http://www.builditsolar.com/Experimental/InWorkshop/SolarShed/solarshed.htm *

*>>*

*>>*

*>> And specifically for the buried pipe here:*

*>> http://www.builditsolar.com/Experimental/InWorkshop/SolarShed/trench.htm *

*>>*

*>> There are some heat loss calcs down toward the bottom.*

*>> My conclusion was that its workable to transfer heat that far, but you *

*>> need to*

*>> have good (thick) insulation, and you should try to do the *

*>> installation in such*

*>> a way as to keep it as dry as possible.*

*>> If you come up with a better scheme, please let us know.*

*>>*

*>>*

--

Gary

www.BuildItSolar.com

gary@BuildItSolar.com

"Build It Yourself" Solar Projects

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> How do you figure out the thermal loss of the pipe?> My thinking is that 100 ft of 3/4" pipe is about 25 square feet of area.> So, 25/R * Temp_Diff = loss in BTU/HR> For R10 at 50 degree that's only 125 BTU/Hr. Calculating R value for> circular vs flat perplexes me... My guess is that the R value would be> lower than what you normally expect per inch.