What Is R Value

Ken-

I won’t attempt to answer what R-value is cold or warm. But, I think I can help with the definition.

The units for R-Value are 1/[Btu/(hr sq ft deg F)].

BTU/hour is rate of heat transfer through the material.

Sq ft is the surface area in square feet of the material.

Deg F is the difference in temperature on each side of the material in Fahrenheit degrees.

So, say you had a 2 ft square piece of material, that passed 50 BTU of heat per hour from one side at 70 degrees to the other side at 40 degrees.

The R-Value would be 1/(50/((2*2)*(70-40))) = R2.4

R-values are convenient to use because they can simply be added together for various layers. E.g. if you put an R2.2 pad on top of an R3.1 pad, you get an R5.3 pad. (This ignores edge effects, air boundary layers, etc.)

Also, for a given material, if you double the thickness, you double its R-Value. (Again, ignoring edge and boundary effects.)

Hope this helps.

Cheers,

-Mike

[edited to correct error found by Bill and Paul in following posts. /mm]

Mike: I once asked Thermarest this question. I wanted to use two of their pads and asked if I could add the “R” value of each for the total, They said no it didn’t work like that. What is your source for this?
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“R-values are convenient to use because they can simply be added together for various layers. E.g. if you put an R2.2 pad on top of an R3.1 pad, you get an R5.3 pad. (This ignores edge effects, air boundary layers, etc.)

Also, for a given material, if you double the thickness, you double its R-Value. (Again, ignoring edge and boundary effects.)” ????
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Thanks

Mike, are you intending to calculate “big R” (Thermal Resistance) or another thermodynamic quantity? Using your formula, double the thickness and see what happens. Did you get the answer (“big R”) that you expected to get?

Essentially, Mike is correct. Keep in mind that as Mike says, certain “effects” are ignored, as is air movement/exchange between the two layers of pads.

Two identical pads can trap a bit of air (which is then warmed) between them which can be pumped out/exchanged with cooler outside air as a sleeping hiker moves about on the pad during the night. This will have some effect on the system as a whole.

Bill, I’ve got to run for a few hours. It’s been over 30yrs since I looked at Heat Transfer equations. I remember them, but I’m very rusty and I’m trying to collect my thoughts so that my post turns out to be shorter than my SMD Essence pack review.

If no one else posts back, I’ll do so later around lunch time. I’ll try to explain a simple equation R = d / k and how Mike’s equation fits into it.

This stuff is basic Heat Transfer (which is good because I only understand it on a basic level). Any undergraduate level Thermodynamics book should cover it.

Bill, if a reply isn’t necessary, just post back and I won’t burden the Forum with another post on this subject. Many thanks, pj

Bill,

First off, please understand that I am not a Thermodynamicist. I just read a textbook many years ago, and don’t remember very much from it. ‘Ok’, let’s see if we can figure this out…

To keep it simple, let’s ignore any units in our formulas. It’s really not necessary to consider them to illustrate how the formula functions in a general sense. Also, let’s assume that we’re considering two identical sized pads (thickness of the pads can be different or the same), or are only concerned with the area of both pads that coincide or are in contact with each other. So if one pad is larger than the other, we’re only considering the area of the smaller pad.

To further keep it simple, let’s just stick to

R = d / k, where:

R is the Thermal Resistance
d = is the thickness of the material
k = the Thermal Conductivity for that material,

k is a constant that could only be coincidentally the same for different types of material. Not being a materials guy, I don’t know if any two materials having the same thickness and area coincidentally just happen to have the same k value when both are subjected to the same temperature differential.

Therefore, to put it a little clearer, k is a constant that is different for each type of material.

Since, for a given material, k is a constant, it will never change for that material. In case you do some research yourself, in some equations a miniscule/lower case greek lambda is used instead of a k; an L for the d – other letter substitutions exist for other equations also.

Therefore, for any given material, the Thermal Resistance, R, is solely dependent upon the value of d (the thickness of the material). Since d is in the numerator, R is directly proportional to d. So, if d increases, so will R.

It’s now easy to see that if d is doubled in value, R will also double in value. You can prove this to yourself by making a simple “test case”. For example, pick values for d and k that would yield a value of,…let’s say…, 10 for R. So, something like d = 20 and k=2. Therefore, 20 divided by 2 = 10, so R would equal 10. Now, double the value of d to 40. This simulates doubling the thickness of the material since d is the thickness of the material. Now doing the arithmetic, R would be 20 and have twice the value if the thickness is doubled.

We could go into further depth regarding how the constant k value known as Thermal Conductivity is calculated, but at this juncture I’m not sure that it is necessary to answer your question.

I hope this info helps.

[Note: If anyone out there reading this would like to comment or correct me on any point, please feel free to help out here. I recall quite a while ago, someone posted an excellent informative post that applied Heat Transfer equations to the heat loss of a human body and how many inches of insulation was required at different temps. Can anyone find it by searching the Forums?]

Paul and Bill-

Thanks for your comments. You caught me! I botched the units in my previous post. The equation I used was for R-Value per unit thickness, which is not what we wanted. I’ll edit my original post later this afternoon with the correct information.

If you combine two thin flat pads, just adding the R-Values will give the correct result. But, if the pads are ridged or really thick, things go astray:

1) On a ridged pad, the boundary air layer created by the ridges may or may not have been included in the claimed R-Value. Further, if you combine two ridged pads, the total insulation depends on how the air spaces line up. E.g. two nested Z-rests will not have twice the R-value of a single Z-rest.

2) As Paul mentioned, with two ridged pads, air can enter from the edges between the pads, reducing the overall R-Value.

3) If the pads are thick compared to their size, relatively more heat is lost through the edges of the pads, so just adding R-Values will overestimate the warmth of the combination.

Cheers,

-Mike