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underground pipe question - Page 2

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


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
 


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


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:


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


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

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


Your welcome


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:

www.BuildItSolar.com
gary@BuildItSolar.com
"Build It Yourself" Solar Projects










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