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You CAN have too much insulation

J

John Gilmer

Jan 1, 1970
0
Hi:

In the E Mail list I subscribe to there was a discussion about insulation.
Someone found a reference that says than when you insulate a cylinder for
any conductivity K of the insulation and a head transfer coefficient H for
transfer of heat from the outside of the insulation to the ambient air when
you increase the insulation beyond some point (dependent upon the ratio of H
& K) then the THICKER the insulation the greater the HEAT LOSS.

I suspect that this is only a practical concern with insulating very hot
tubes. But it was an interesting result.
 
R

Robert Morein

Jan 1, 1970
0
John Gilmer said:
Hi:

In the E Mail list I subscribe to there was a discussion about insulation.
Someone found a reference that says than when you insulate a cylinder for
any conductivity K of the insulation and a head transfer coefficient H for
transfer of heat from the outside of the insulation to the ambient air when
you increase the insulation beyond some point (dependent upon the ratio of H
& K) then the THICKER the insulation the greater the HEAT LOSS.

I suspect that this is only a practical concern with insulating very hot
tubes. But it was an interesting result.
This is intuitively counterintuitive, and I wonder if something has been
left out of the discussion.
The Heat Equation has only dissipative terms. Although the solution is
composed of Fourier terms, there are no local maxima.
The heat flux through the head can only be proportional to the temperature
differential, and no more.
It's not like the Wave Equation, where if you increase the damping term past
the critical value, the settling time actually increases.
 
N

no useful info

Jan 1, 1970
0
Robert Morein said:
This is intuitively counterintuitive, and I wonder if something has been
left out of the discussion.
The Heat Equation has only dissipative terms. Although the solution is
composed of Fourier terms, there are no local maxima.
The heat flux through the head can only be proportional to the temperature
differential, and no more.
It's not like the Wave Equation, where if you increase the damping term past
the critical value, the settling time actually increases.

aw if you only had intuition. you are so easily side-tracked
 
John Gilmer said:
...Someone found a reference that says than when you insulate a cylinder for
any conductivity K of the insulation and a head transfer coefficient H for
transfer of heat from the outside of the insulation to the ambient air...

Sounds like h is the surface airfilm thermal conductance, ie the heat
(vs head) transfer coefficient, about 1.5 Btu/h-F-ft^2 for slow-moving
air. R5 per inch Styrofoam has k = 1/(12x5) = 1/60 Btu/h-ft-F.
when you increase the insulation beyond some point (dependent upon the
ratio of H & K) then the THICKER the insulation the greater the HEAT LOSS.

Not exactly. The cylinder's thermal conductance (2Pi/(ln(r/ri)/k+1/(hr))
Btu/h-F-ft) is maximized at the critical radius Rcrit = k/h = 1/(1.5x60)
= 0.0111' or 0.133 inches for styrofoam. That's the outer radius, including
the insulation. Up to that radius (about 1/4" in this case), adding more
Styrofoam increases the heat loss. Above it, adding more decreases the
heat loss.

Read all about it on page 25 of the 1998 Schaum's Outline on Heat Loss.

Nick
 
D

daestrom

Jan 1, 1970
0
John Gilmer said:
Hi:

In the E Mail list I subscribe to there was a discussion about insulation.
Someone found a reference that says than when you insulate a cylinder for
any conductivity K of the insulation and a head transfer coefficient H for
transfer of heat from the outside of the insulation to the ambient air when
you increase the insulation beyond some point (dependent upon the ratio of H
& K) then the THICKER the insulation the greater the HEAT LOSS.

I suspect that this is only a practical concern with insulating very hot
tubes. But it was an interesting result.

I think what you're group is thinking of is known as 'critical insulation
thickness'. When looking at very small diameter cylinders such as heated
wires, or small pipes, this problem comes up.

As insulation is added to the pipe/cylinder, the outer surface area
increases. With very small pipes (or very poor insulating materials), the
increase in surface area can increase the heat losses faster than the
increase in thickness of the insulation can decrease the heat loss. So as
insulation is added, the heat losses get worse. But once the diameter
reaces the 'critical insulation thickness', the surface area of the cylinder
wall does *not* increase the heat losses as fast as the thicker insulation
decreases it.

The critical radius is given by R = lambda / h0 where lambda is the
thermal conductivity. For most pipes and wires in still air, the critical
radius is on the order of 2.5 cm or less depending on the exact insulation
used. If the overall radius is larger than the critical radius, then the
heat losses decrease as more insulation is added. So if you think you're in
that realm, just add more insulation to get beyond the critical radius.

It is only a problem with shapes whose outer surface area grows rapidly when
a thin layer is added. Flat walls and such it doesn't become a problem.

daestrom
 
D

daestrom

Jan 1, 1970
0
Robert Morein said:
of
This is intuitively counterintuitive, and I wonder if something has been
left out of the discussion.
The Heat Equation has only dissipative terms. Although the solution is
composed of Fourier terms, there are no local maxima.
The heat flux through the head can only be proportional to the temperature
differential, and no more.
It's not like the Wave Equation, where if you increase the damping term past
the critical value, the settling time actually increases.

For cylinders, the area of the exterior is the term to look at. As a layer
of insulation is added, the surface area through which the heat is conducted
increases. If it increases faster than the insulating effects of extra
insulation, the losses increase.

daestrom
 
R

Robert Morein

Jan 1, 1970
0
daestrom said:
ratio

For cylinders, the area of the exterior is the term to look at. As a layer
of insulation is added, the surface area through which the heat is conducted
increases. If it increases faster than the insulating effects of extra
insulation, the losses increase.

daestrom
Thank you. Makes perfect sense.
 
A

Anthony Matonak

Jan 1, 1970
0
daestrom wrote:
....
For cylinders, the area of the exterior is the term to look at. As a layer
of insulation is added, the surface area through which the heat is conducted
increases. If it increases faster than the insulating effects of extra
insulation, the losses increase.

I guess the trick is to use insulation which does not conduct heat
very well. :)

Anthony
 
R

Robert Morein

Jan 1, 1970
0
Anthony Matonak said:
daestrom wrote:
...

I guess the trick is to use insulation which does not conduct heat
very well. :)

Anthony

What a novel idea! :).
 
B

Bob Adkins

Jan 1, 1970
0
Hi:

In the E Mail list I subscribe to there was a discussion about insulation.
Someone found a reference that says than when you insulate a cylinder for
any conductivity K of the insulation and a head transfer coefficient H for
transfer of heat from the outside of the insulation to the ambient air when
you increase the insulation beyond some point (dependent upon the ratio of H
& K) then the THICKER the insulation the greater the HEAT LOSS.

I suspect this was due to a high temperature differential, and all the thick
insulation soaked up a bit of heat.

In doing such an experiment, the insulated compartment and insulation should
be allowed to normalize before measurements are taken.

Bob
 
F

Fred B. McGalliard

Jan 1, 1970
0
....
For cylinders, the area of the exterior is the term to look at. As a layer
of insulation is added, the surface area through which the heat is conducted
increases. If it increases faster than the insulating effects of extra
insulation, the losses increase.

But, given the internal pipe plus insulation, adding a layer of external
insulation decreases the net heat loss, but by much less than the internal
layer because the area is larger. Paying more for less improvement, but you
never get to paying more for nothing. That's clear isn't it?
 
F

Fred B. McGalliard

Jan 1, 1970
0
....
Not exactly. The cylinder's thermal conductance (2Pi/(ln(r/ri)/k+1/(hr))
Btu/h-F-ft) is maximized at the critical radius Rcrit = k/h = 1/(1.5x60)
= 0.0111' or 0.133 inches for styrofoam. That's the outer radius, including
the insulation. Up to that radius (about 1/4" in this case), adding more
Styrofoam increases the heat loss. Above it, adding more decreases the
heat loss.

This must be presuming that the heat coupling from the outside of the
insulation to the still air is, for very narrow pipes over some temperature
range, limited by the external conditions. If you blow the air over the
surface, or put it under water, or provide a reasonable amount of insulation
so the external surface temperature is driven by the internal heat flow and
not the coupling, then our "common sense" expectations must be met, right?
 
D

daestrom

Jan 1, 1970
0
Fred B. McGalliard said:
...

But, given the internal pipe plus insulation, adding a layer of external
insulation decreases the net heat loss, but by much less than the internal
layer because the area is larger. Paying more for less improvement, but you
never get to paying more for nothing. That's clear isn't it?

Let's take an 'extreme' example:

You have a wire that is 2mm in diameter. And we'll assume the
free-convective heat-transfer coefficient with air is 20 W/m^2-K, and the
wire surface is 75C above the local air temperature.

For a length (L) of the wire, the surface area is pi*(0.002) * L = 6.28e-3 *
L m^2 so the heat loss per unit length is 20 * 6.28e-3 * L = 1.26e-1 * L
watts.

Now, lets put a 2mm coating of insulation that has a conductivity of 0.035
W/m-K (about the same as fiber-glass). The surface area of the outside of
the insulation is pi*(.006) * L = 1.885e-2 * L m^2. But we don't know the
temperature at the outer surface of the insulation though, so lets just say:
Q = 1.885e-2 * L * 20 *(Tsurface - Tambient) watts
= 3.77e-1 * L * (Tsurface - Tambient) watts.

Similarly, we know the inside temperature of the insulation and the heat
transfer through such a cylinder is
Q = (Twire - Tsurface) * 2*pi* 0.035 * L / ln(.003/.001) (.003 is the outer
radius of the insulation, .001 the inner radius)
= 2.002e-1 * L * (Twire - Tsurface) watts.

The heat loss through the insulation must be equal to the heat lost through
the air film when in the steady-state, so setting the two equations equal to
one another...
3.77e-1 * L *(Tsurface - Tambient) = 2.002e-1 * L * (Twire - Tsurface)

The temperature drop across the insulation is 1.88 times the temperature
drop across the film. If the total is 75C, then the temperature drop across
the insulation is 3.77e-1 / (2.002e-1 + 3.77e-1) * 75 = 49C. Across the air
film (75-49)=26 -or- 49/1.88 = 26

Now we go back to see what the Q is for the insulation = 2.002e-1 * L * 49C
= 9.8 * L watts. Quite a bit more than the bare wire alone (at 0.126 * L
watts). (a check is to calculate the Q for the air film as 3.77e-1 * L *
(75-49) = 9.8 * L watts)

So in this 'extreme' case, putting the insulation on *increased* the heat
loss from 1.26e-1 watts / meter to 9.8 watts / meter. (paying for making
it worse).

The problem doesn't come up in 'ordinary' applications in homes. If the
initial radius of the pipe is larger, then the calculations work out the way
most people expect. But with *tiny* wires and pipes, it can be a surprise
to actually make the heat loss *worse* by adding insulation.

daestrom
 
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