# How do you know you can add thermal resistance but not thermal transmittance?

I followed through the example on this wikipedia page for calculating the thermal resistance of a composite material (a wall) composed of many layers: https://en.wikipedia.org/wiki/Thermal_transmittance#Calculating_thermal_transmittance

Thickness Material Conductivity Resistance = thickness / conductivity
Outside surface 0.04 K⋅m2/W
0.10 m (0.33 ft) Clay bricks 0.77 W/(m⋅K) 0.13 K⋅m2/W
0.05 m (0.16 ft) Glasswool 0.04 W/(m⋅K) 1.25 K⋅m2/W
0.10 m (0.33 ft) Concrete blocks 1.13 W/(m⋅K) 0.09 K⋅m2/W
Inside surface 0.13 K⋅m2/W

[...the] total resistance is 1.64 K⋅m2/W [...] thermal transmittance [is] 0.61 W/(m2⋅K).

I understand intuitively that you can not add the thermal transmittances together to get the final result as this "does not make sense". But from a purely "dimensional analysis" point of view it is perfectly reasonable to add numbers of the same units (in this case W/m2⋅K). So I assume there is not some mathematical rule or heuristic that I can reference and instead there must be some physics rule or heuristic that I can reference? I almost very easily made this mistake and would like to know if there is some simple way to avoid making a similar mistake with other physical systems?

Or is it just an inherent part of physics as mathematically you can do many things that do not make physical sense and physics is the deliberate systematic act of observing and experimenting to limit down the number of "allowed" mathematical operations so as to only perform calculatings that add value / cohere with reality? Perhaps I should post this to philosophy.stackexchange.com instead.

• Why can you add resistors in series, but not capacitors? And if you can add resistance, why not conductance? Same kind of thing... Commented Aug 17, 2022 at 16:52
• Thank you @JonCuster.
– AJP
Commented Aug 17, 2022 at 17:21

From this, we find that if two objects are placed in end-to-end contact, then any interface point has a single temperature, that energy entering (leaving) from the left side must equal the energy leaving (entering) from the right side, and that this energy flow must be $$q=k_iA\frac{\Delta T_i}{L_i}=A\frac{\Delta T_i}{R_i}$$ for object $$i$$, where $$k$$ is the thermal conductivity, $$A$$ is the cross-sectional area, $$\Delta T$$ is the temperature difference down the length $$L$$ of the object, and $$R$$ is the thermal resistance.
$$T_\text{interface}=\frac{\sum_i(k_iT_i/L_i)}{\sum_i (k_i/L_i)},$$
and so the heat flow is $$q=A\frac{T_1-T_2}{\sum_i R_i}$$, which holds generally for even $$i>2$$ if the end temperatures are taken as $$T_1$$ and $$T_2$$. Note that the thermal resistances (not the thermal conductances $$1/R_i$$) end up adding.