Can someone explain K factor ?

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BJN644

Feeling the Heat
Hearth Supporter
Nov 4, 2007
352
Maine
I have stove that requires floor protection with a K factor of .84, can anyone translate that into layman terms ?

Thanks
 
BJN644 said:
I have stove that requires floor protection with a K factor of .84, can anyone translate that into layman terms ?

Thanks


k = thickness in inches/R

so, solving for "R" and we get

R = thickness/k

Example. Micore 300 I think has a K value of 0.43 so, to compute the R-value of 1 inch of it we get this


R = thickness/k

R = 1/0.43

R= 2.33

saying you need a K-value alone is not sufficient because the R-value is determined by the thickness AND the K-value. Are you sure it doesn't say something like 1-inch of material with a K-value of 0.43 is required or some other combination of thickness and K-value? If you can't find that, tell us the stove make and model and we'll look at the manual......it should give the above or an R-value
 
this is a calculation you may be able to use, i pulled it from one of our manuals. hope this helps

An easy means of determining if a proposed alternate floor protector meets requirements is to follow this procedure:
1) Convert specification to R-value:
i R-value is given – no conversion is needed
ii k-factor is given with a required thickness (T) in inches: R = 1/k x T
iii C-factor is given: R = 1/C
2) Determine the R-value of the proposed alternate floor protector:
i Use the correct formula given in step 1 (above) to convert values not expressed as “R.”
ii For multiple layers, add R-values of each layer to determine overall R-value.
3) If the overall R-value of the system is greater than the R-value of the specified floor protector, the alternate is acceptable.

EXAMPLE:
The specified floor protector should be ¾” thick material with a k-factor of 0.84. The proposed alternate is 4” brick with a C-factor of 1.25 over 1/8” mineral board with a k-factor of 0.29.

Step (a): Use formula above to convert specification to R-value.
R = 1/k x T = 1/0.84 x .75 = 0.893
Step (b): Calculate R of proposed system.
4” brick of C = 1.25, therefore R brick = 1/C =1/1.25 = 0.80
1/8” mineral board of k = 0.29, therefore Rmin.bd. = 1/0.29 x 0.125 = 0.431
Total R = Rbrick + Rmineral board = 0.8 + 0.431 = 1.231
Step (c): Compare proposed system of R of 1.231 to specified R of 0.893. Since proposed system R is greater than required, the system is acceptable.
Definitions:
Thermal conductance = C = ____Btu_____ = ____W____
(hr)(ft2)(deg F) (m2)(deg K)

Thermal conductivity = k = __(Btu)(inch)__ = ____W___ = ____Btu____
(hr)(ft2)(deg F) (m)(deg K) (hr)(ft)(deg F)

Thermal resistance = R = (ft2)(hr)(deg F) = (m2)(deg K)
Btu W
 
castiron said:
BJN644 said:
I have stove that requires floor protection with a K factor of .84, can anyone translate that into layman terms ?

Thanks


k = thickness in inches/R

so, solving for "R" and we get

R = thickness/k

Example. Micore 300 I think has a K value of 0.43 so, to compute the R-value of 1 inch of it we get this


R = thickness/k

R = 1/0.43

R= 2.33

saying you need a K-value alone is not sufficient because the R-value is determined by the thickness AND the K-value. Are you sure it doesn't say something like 1-inch of material with a K-value of 0.43 is required or some other combination of thickness and K-value? If you can't find that, tell us the stove make and model and we'll look at the manual......it should give the above or an R-value

BJN644 and I have the same stove, the Harman Oakwood. I tried to attach a scan of the page from the manual but can't. Anyway it's on page 11 of the manual.

Here is a link to the manual: (broken link removed to http://www.harmanstoves.com/doc/oakwoodm.pdf)
 
MANIAC said:
castiron said:
BJN644 said:
I have stove that requires floor protection with a K factor of .84, can anyone translate that into layman terms ?

Thanks


k = thickness in inches/R

so, solving for "R" and we get

R = thickness/k

Example. Micore 300 I think has a K value of 0.43 so, to compute the R-value of 1 inch of it we get this


R = thickness/k

R = 1/0.43

R= 2.33

saying you need a K-value alone is not sufficient because the R-value is determined by the thickness AND the K-value. Are you sure it doesn't say something like 1-inch of material with a K-value of 0.43 is required or some other combination of thickness and K-value? If you can't find that, tell us the stove make and model and we'll look at the manual......it should give the above or an R-value

BJN644 and I have the same stove, the Harman Oakwood. I tried to attach a scan of the page from the manual but can't. Anyway it's on page 11 of the manual.

Here is a link to the manual: (broken link removed to http://www.harmanstoves.com/doc/oakwoodm.pdf)

That's the one.
 
Trying to sort through all the complications, the key is that a K factor is always in inches, whereas an R factor can be given for any thickness?

The other little tidbit is that R factors tend to be used for building materials at normal room and house temperatures, so K seems to be more used in the engineering community and if often specified for a particular temperature range.

As Mike says, something with a K factor of 1 is the same an R factor of 1 (per inch), whereas something with a K factor of .5 is R of 2, and something with a K factor of 2 would be r=.5 - in other words, K x R should equal one.

UL should not (in my opinion) allow these formula in the owners manual, but should require the K and R values plus 3 examples of materials that meet them.
 
Webmaster said:
Trying to sort through all the complications, the key is that a K factor is always in inches, whereas an R factor can be given for any thickness?

The other little tidbit is that R factors tend to be used for building materials at normal room and house temperatures, so K seems to be more used in the engineering community and if often specified for a particular temperature range.

As Mike says, something with a K factor of 1 is the same an R factor of 1 (per inch), whereas something with a K factor of .5 is R of 2, and something with a K factor of 2 would be r=.5 - in other words, K x R should equal one.

UL should not (in my opinion) allow these formula in the owners manual, but should require the K and R values plus 3 examples of materials that meet them.

R is a value that can be in any thickness, but the thickness must be specified as part of the description - IE Durock is R-0.26 for a 1/2" layer.

I don't agree with you on not putting the formulas in the manuals, if anything I think that UL should REQUIRE the formulas, ACTUAL VALUES, (or that a simply non-combustible surface is all that's needed) AND ALSO the three examples of materials (or combinations) that comply, with the proviso that the materials need to be commonly available "standard" building products, or at least ones that can be specifically identified enough to tell what their values are. I've seen several threads where people were trying to figure out what the specified material was supposed to be, so they could determine what R-value was needed in an equivalent.

Gooserider
 
This is what the manual says, again in laymen terms what is required ?

The Oakwood must have a floor protector with a
k-value of .84 or higher to meet UL safety standards (unless
purchased with optional bottom shield).
In all installations, the area under and around the stove
must be protected from falling ash and live coals. The area
under a horizontal run of chimney connector must also
be protected . This protector must be of noncombustible
material, and positioned as shown in the accompanying illustration.
The guidelines for floor protection are as follows:
In the U.S. the floor protector must be completely
under the stove, 16" in front of the door opening, 8" to each
side and 8" to the back. If there are any horizontal runs of
the stove pipe the ash protector must extend 2" on each
side of the pipes shadow.
In Canada, the ash protector must extend 18" in front
of the door opening, 8" to each side and extend to the wall
behind the stove.
Clearances
 
The manual quote you give simply can't be translated into anything useful as written. The problem is that they say "k-value of .84 or higher" which is a meaningless statement because the "k-factor" is a measurement of heat flow PER INCH - thus the quote is sort of like saying "I'm traveling at a speed of 10" - this does not say anything useful because you don't say what the unit is.

If one makes the gross assumption that they are talking about the K-value of a one inch thick pad, which doesn't seem totally unreasonable in this context, then you are looking at a moderate insulation requirement, per the formula, 1/.84 x 1" gives an R-value requirement of 1.19

Probably the thinnest pad would be 1/2" of Micore, (R-1.1) topped with a layer of 1/2" Durock (R-0.26) and tile (negligible). The easiest would be 5 layers of 1/2" Durock. If you look at the Wiki article on hearth construction, there is a table of common non-combustible materials and their R-values. Any combination that is structurally sound, and that adds up to a TOTAL equal to, or greater than the stated requirement will work.

The problem though, gets back to translating the manual. Since Harman is not going to want to talk to you directly, you need to go to your local dealer with the manual, and ask him what the requirement is. If all else fails, get him to show you a commercial pad that is approved for that stove and get the value from it.

Gooserider
 
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