Lets say you have an R-value of 15. The top side is 35C, the bottom
side is 75C. What would be the rate of transfer? How would that change
if the sides were flipped? What if it was a verticle wall? Please
explain each variable and show the formula, as well as an example
which uses the formula.

I have seen at least one formula on the internet, but it does not
sound right to me because I would expect that the larger the
temperature difference, the exponentially larger the heat transfer
will be. However, the equation did not seem to incorporate that in the
formula.

General formula for flow

  flow =  Delta  * area * conductivity / thickness

applies to many, if not all, flows, be it heat, electricity or fluid.

In case of heat, Delta is difference of temperatures on the two sides
of block of material
and conductivity is thermal conductivity k - a property of the material.

 The product:  area * conductivity / thickness 
 is called conductance and is inverse of resistance.

 The ratio: Delta / thickness is called gradient - in case of heat -
the temperature gradient.

So - you are right that flow is more when Temperature difference is larger,
but not exponentialy more, only lineraly more. It also is proportional to the
area. So these two numbers would have to be suplied to do the calculation.
The property is symetrical - meaning : if you exhabe the sides, hot for cold,
you get the same flow (in the new direction) but you have to wait
after a steady flow is established.

The values of conductivity, and proper SI units are detailed here:

http://en.wikipedia.org/wiki/Thermal_conductivity


R- value is used in construction industry in US and uses not the SI units,
but imperial units of measurements (BTUs inches and feet).

CONVERSION:

 SI unit of thermal resistance = K·m²/W, 
 Imperial unit is = 1 ft² F° h / Btu = approx. 0.1761 K·m²/W.

as explained (and converted) here:
http://en.wikipedia.org/wiki/R-value


In the past there was a lot of controversy and arguments about proper labelling,
even lobyying, and false advertising claims concerning R-labels. 
Today,R-value is defined by a specific ASTM test which measures heat
flow  in BTUs per hour,
through  1 feet square, one inch thick material.

http://www.epsmolders.org/PDF_FILES/RVALUES.PDF

Hedgie

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Request for Answer Clarification by centure7-ga on 13 May 2005 14:05 PDT

Hi Hedgie, thanks for your response, but I'm still not sure about a
couple of things:

1. Lets say you have a hot insulated body suspended in the middle of
some air. Even if the R-value on the top of that body is identicle to
the R-value at the bottom, you will get much more heat loss through
the top, won't you? Since heat rises, I would think much more heat
would be lost from the top surface, despite the identicle R-value.
Looking at the "reflective insulation" section at
http://www.mge.com/business/Madison/PA_45.html, it seems to me to be
saying that you get a different R-value based on the same material in
different situations, though the wording is not clear to me.

2. I see from the website I provided in the last question that to
convert to a conductivity value, you can use the formula 1/R to
determine the U-value, which is Btu/square foot-°Fahrenheit-hour.  Is
that the value I would plug in as the conductivity value in the flow
equation? Or, I see you have mentioned a conversion formula to convert
Btu/square foot-°Fahrenheit-hour to K·m²/W (u-value * 0.1761).

---

Clarification of Answer by hedgie-ga on 13 May 2005 22:58 PDT

centure7-ga

re 1) The article you mention 
http://www.mge.com/business/Madison/PA_45.html
is good, except for last paragraph
 "Reflective insulation"
which is an attempt to extrapolate the R concept beyond its domain of validity.

The physics topic is 'Heat transfer' which happens by conduction,
radiation and  convection
http://en.wikipedia.org/wiki/Heat_transfer

R-values deal with conduction and depend on the whole insulation block
of given thickness
 - you cannot have different value on the top or bottom
 - and all heat 'comming in at one end' must 'come out' on the other end
     (since we are in 'steady state')

Reflectivity is a surface property (surface painted black or white)
and it should not be confused with R values. Radiant heat (infra-red
light) is either
absorbed in the surface (increasing its Temperature) or reflected. Increased
temperature will lead to increase of conductive flow.

Both effects can be measured if there is no other transfer going on:
If you hang a block of material in the room, dominant mechanism will
be convection - air flowing around the block will bring surfaces on
both sides to same temperature. If you have a hot (or cold) stuff inside -
like in a cooler - then T1 and T2 is temperature in the cooler and temperature
in the room and flow is either from inside to outside or the opposite to that.

So, to measure other modes, you need to eliminate convection: Have two
rooms (or boxes) well insulated, adjacent, with a window between them.
One room is heated to temp T1 , other to T2 and window is filled with
a block of material.
Then R value of that block will rule the flow of heat according to the
Flw Equation above.

So, radiant barriers differ from resistive insulation 
http://www.fsec.ucf.edu/pubs/energynotes/en-15.htm

If both rooms are in thermal equilibrium, radiant heat will not 
influence measurement of the R-value. To see effect of Radiant Heat,
add a 'heat-lamp' to shine on one side of the window. Depending on
reflectance of that surface, the flow of heat will increase.

re 2) 
We need to differentiate between conductivity and conductance


which are  either property of

material      or of a     block with thickness T
------------                    ----------
conductivity                   conductance
resistivity                    resistance

Vikipedia article
http://en.wikipedia.org/wiki/Thermal_conductivity

says correctly:

in summary, for a plate of thermal conductivity ?, area A and thickness T:

thermal conductance = ?/T, measured in W/K·m^2
thermal resistance = T/?, measured in K·m^2/W

but then  confuses the issue by a  comment in parenthesis.

 In the flow formula you have 

flow =  Delta  * Area * conductivity / thickness
is same as
flow = Delta * Area * conductance

  as conductance = ?/T = conductivity / thickness

3) In conclusion

overview of all modes is here
http://www.radiantbarrier.com/physics_of_foil.htm

Hedgie

---

Clarification of Answer by hedgie-ga on 13 May 2005 23:15 PDT

One more link - your tax money at work :-)

 National labs have been directed to help citizens to save energy

Hre is page which points to some results:

http://hes.lbl.gov/hes/glossary.html

Click on
Insulation Terminology
 ...... and it looks like nothing happened, you scroll down and find

Insulation Terminology

Insulation 
with links to 

Department of Energy's Insulation Fact Sheet, produced by Oak Ridge
National Laboratory

Radiant barriers is the Department of Energy's Radiant Barrier Fact
Sheet, produced by Oak Ridge National

etc.

---

Request for Answer Clarification by centure7-ga on 28 May 2005 02:41 PDT

Before I rate your answer, please provide an example calculation using
the variables indicated in the original question (40C difference,
R-15).

---

Clarification of Answer by hedgie-ga on 31 May 2005 04:54 PDT

If you want to measure flow across the  whole block,
you need to know the area of the block.

Let's assume you work in the Customary (aka British Imperial) units 
and area is 4 ft^2 (four square feet), then flow is

Flow=A * Delta * conductance = 4 * delta /R = 4 * 40 /15 

 I ype 4*40/15 into google and get:
(4 * 40) / 15 = 10.6666667 

that is your answer in Btu/hour

 If Your area is A in feet square, your can use formua

Flow= A *8 */3 Btu/hour

Additional example and details are in 
 
http://en.wikipedia.org/wiki/R-value

where it says:
Some countries use a non-SI definition: R = ft^2 F° h/Btu.
The conversion between the two is is 1 ft² F° h / Btu = approx. 0.1761 K·m²/W.

 For a practical example, if the interior of your home is at 20°C, and
the roof cavity is at 10°C, that gives a temperature difference of
10°. Then, assuming a ceiling insulated to R2, 5 Watts of energy will
be lost for every square metre of ceiling. This is of course a
theoretical example, and therefore fails to take into account a myriad
of other factors.

Hedgie

Thanks for going out of your way to answer the question completely thoroughly.

Hedgie-ga?s answer is correct, however it is not generally appreciated
that the thermal conductivity (R value) is temperature dependant. For
most construction applications the difference is small and can be
ignored. It becomes important in designing the thickness of furnace
linings. A formula that works well for most common materials is: lnk =
a + bT, where:

lnk is the natural logarithm of the thermal conductivity
T is the absolute temperature (°C + 273)

Ie The natural log of the thermal conductivity is a straight line
where ?b? is the gradient of the slope and ?a? is the intercept on the
lnk axis.

If you know the thermal conductivity at two different temperatures
then you can use quadratic equations to eliminate ?b? and find ?a?
The value of ?a? can then substituted in either equation to find ?b?
Knowing ?a? & ?b? the thermal conductivity at any other temperature can be found.

lnk1 = a + bT1
lnk2 = a + bT2

In applying the formula to a real situation it is the ?k? value at the
higher temperature that is generally used. In your case 75 + 273
=348°K.
This relationship works well for most common lightweight materials
such as fibre and thermolite brick but becomes less accurate for very
dense materials.

Hope this helps. 

Norman

If you want to measure flow across the  whole block,
you need to know the area of the block.

Let's assume you work in the Customary (aka British Imperial) units 
and area is 4 ft^2 (four square feet), then flow is

Flow=A * Delta * conductance = 4 * delta /R = 4 * 40 /15 

 I ype 4*40/15 into google and get:
(4 * 40) / 15 = 10.6666667 

that is your answer in Btu/hour

 If Your area is A in feet square, your can use formua

Flow= A *8 */3 Btu/hour

Additional example and details are in 
 
http://en.wikipedia.org/wiki/R-value

where it says:
Some countries use a non-SI definition: R = ft^2 F° h/Btu.
The conversion between the two is is 1 ft² F° h / Btu = approx. 0.1761 K·m²/W.

 For a practical example, if the interior of your home is at 20°C, and
the roof cavity is at 10°C, that gives a temperature difference of
10°. Then, assuming a ceiling insulated to R2, 5 Watts of energy will
be lost for every square metre of ceiling. This is of course a
theoretical example, and therefore fails to take into account a myriad
of other factors.

Hedgie

PS - Regarding my rating comment, I now realize that "completely
thoroughly" is an obvious tautology!