Can R values be averaged?

Question

Suppose there is a wall and half of its area has R-value 10, and the other half has R-value 20. Is the total R-value of the wall 15? In other words, can R-values be averaged weighted by area?

Please justify your answer with math. R-value is defined as follows on page 454 of the 2020 Residential Code of NY State.

R-VALUE (THERMAL RESISTANCE). The inverse of the time rate of heat flow through a body from one of its bounding surfaces to the other surface for a unit temperature difference between the two surfaces, under steady state conditions, per unit area (h • ft2 • °F/Btu) [(m2 • K)/W].

Answer 1

No, R values can't be averaged to find an assembly's effective R value. R value is a local material property, thickness divided by thermal conductivity, and the physics don't allow for simple averaging.

Given a steady state temperature difference ΔT between an insulation assembly's interior and exterior, the heat flux (W/m²) is ΔT/R. Fixing two assemblies with unique areas and R values, the heat transfer rate for each assembly is A₁·ΔT/R₁ and A₂·ΔT/R₂. To compute an effective R value, I want the R₁₂ that combines with A₁+A₂ to yield the same heat transfer rate as both assemblies combined, i.e.

A₁·ΔT/R₁ + A₂·ΔT/R₂ = (A₁+A₂)·ΔT/R₁₂.

Solving for R₁₂, I get

R₁₂ = (A₁+A₂) / (A₁/R₁ + A₂/R₂).

Note that this is not simply the weighted average of R₁ and R₂.

Your hypothetical wall assembly with half R10 and half R20 has an effective R value of 2/(1/10+1/20) = 13.3. When your building code calls for R20 (or whatever), however, it is understood that thermal bridging by wall studs etc. will reduce the effective R factor below the labeled value on your insulation, and that's okay.

A cleaner alternative to R values:

By introducing U = 1/R, area weighted averaging does work for computing effective U values. Recalling

A₁·ΔT/R₁ + A₂·ΔT/R₂ = (A₁+A₂)·ΔT/R₁₂,

eliminating ΔT, and substituting U values, I get A₁·U₁ + A₂·U₂ = (A₁+A₂)·U₁₂ and therefore

U₁₂ = (A₁·U₁ + A₂·U₂) / (A₁+A₂).

For effective insulation values in the OP's case, then, these U values are much easier to use than R values. U values aren't perfectly superior to R values, however. If, for instance, you were to stack R15 and R19 batts, you can add the R values to determine that the assembly has an R value of 34. For U values you would need a 1/(1/U₁ + 1/U₂) computation.

Answer 2

Consider a flat panel shape, thickness E, area S. For example, a wall.

For the purposes of this calculation we consider thickness to be small compared to the two other dimensions so edge effects are not taken into account.

DeltaT is the temperature difference on both sides of the wall.

P is the power going through the wall (thermal loss).

The meaning of terms Resistance and Conductance are identical to their electrical equivalent:

Thermal conductance Gth = P/DeltaT in W/°K

Thermal resistance Rth = DeltaT/P in °K/W.

(Conductance is G for electricity and U for thermal stuff, I went with G).

These are both defined for the whole wall. For example a wall with thermal conductance 10W/°K (thermal resistance 0.1°K/W) will lose a power of 10W per every °K of DeltaT. That's for the whole wall, not "per unit area".

In a somewhat misleading misuse of vocabulary, "R-value" is sometimes called "resistance", while it actually means "Thermal resistance of a wall having a surface equal to one unit area".

If our wall has a certain Rvalue, it means each unit area (1m²) has a Rth equal to the Rvalue. Thermal losses of each m² of wall are DeltaT/Rvalue, therefore the whole wall of area S has losses P=S*DeltaT/Rvalue, which means Rth for the whole wall is Rvalue/S. Larger wall = higher losses = lower Rth.

If you want to mix different insulations, it is much easier to use conductance instead.

Conductance is simply the inverse of resistance, so a wall of unit area (1m²) has thermal conductance 1/Rvalue, and a wall of area S has S times more thermal conductance. For the whole wall, Gth=S/Rvalue. Losses are P=DeltaT*Gth.

It's the same thing with electricity: when several conductances are in parallel, the result is the sum of all conductance values.

So it is correct to speak of "conductance per m²" (or "cost per m²") because "per" means the total is divided by the area. "Thermal resistance per m²" doesn't mean anything, because the division is the other way around.

If we have two walls of different area S1,S2 and Rvalue1,Rvalue2 their losses add up. The conductance of each wall is Gth1=S1/Rvalue1 and Gth2=S2/Rvalue2, which gives a total conductance Gth=Gth1+Gth2.

Total losses are P=DeltaT*Gth.

If you use resistances, the formula is the same as parallel resistances in electricity, taking the area into account:

Rth = 1/Gth = 1/( S1/Rvalue1 + S2/Rvalue2 )

"Rvalue for the whole wall" = (S1+S2)/( S1/Rvalue1 + S2/Rvalue2 )

Just like a short circuit in parallel with anything is still a short circuit, the dominant part of this is the area with worst insulation. Basically, if half the wall has R-value 5, and the other half of the wall has R-value 1, the resulting R-value of the whole wall is not the average (ie 3) but rather 1.666...

Note it makes no sense to define R-value for a wall that has different insulation on one half and the other half, because R-value is a property of a homogenous material. A more correct way to say it would be "losses would be the same if the whole wall was insulated with R-value 1.6".

If some readers don't like math, think of cost instead. Thermal loss per m² and conductance are just like cost per m². It is correct to make weighted average (by area) of cost or conductance, because these things add, the total cost is cost per m² multiplied by area. To keep the cost analogy, resistance is like m²/€, you can't add nor average that.