# B Thermal resistance without area

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1. Apr 25, 2017

### whatdoido

I'm (self)studying the physics of heat transfer at the moment. My book gives a relationship between heat transfer rate and thermal resistance as $\phi=\frac {A \Delta T} {R}$. My book is not in English, so hopefully that is not the cause of this misunderstanding. I double checked that heat flow rate means the same as in my native language. Heat flow rate can be also marked as $q$ and $Q$ I think since I saw both being used in different places.

When I read about thermal resistance from other sources in the internet I ran across this equation $\phi=\frac {\Delta T} {R}$. I give a specific source: http://web2.clarkson.edu/projects/subramanian/ch330/notes/Conduction in the Cylindrical Geometry.pdf

The heat flow rate of cylinder is defined the same as in my book, but the relationship between heat flow rate and resistance differ, since it does not include area in it. Well maybe I mixed heat flow rate with heat flux? Then it would not make sense that cylinder's heat flow rate matches with the one in my book. Something I'm not understanding correctly, so any help would be much appreciated.

2. Apr 25, 2017

### Staff: Mentor

If the equation doesn't involve the area, then it gives heat flow rate per unit area (also known as heat flux). If it does involve the area, then it is total heat flow rate.

3. Apr 26, 2017

### whatdoido

In that source the heat flow rate of cylinder is $Q=2\pi L k \frac {T_1 - T_2} {ln(r_2 / r_1)}$. Since the symbol $Q$ is reserved for heat flow rate, then how can $Q=\frac {\Delta T}{R}$ make sense? It does not have area in it, so $Q$ should mean heat flux. See my confusion? It is even mentioned that $Q$ means heat flow rate.

4. Apr 26, 2017

### Staff: Mentor

Strictly speaking, the relationship in your original post only always works if the equation is expressed in differential form: $$q=-k\frac{dT}{dx}$$ and $$Q=-Ak\frac{dT}{dx}$$. In the latter equation, A might even be a function of x. Here's how it applies to the equation you have written in this post (for heat flow through a pipe wall):
$$q--k\frac{dT}{dr}$$
and $$Q=-(2\pi rL)k\frac{dT}{dr}\tag{1}$$
Eqn. 1 applies to every radial location between r1 and r2.

The equation $Q=2\pi L k \frac {T_1 - T_2} {ln(r_2 / r_1)}$ is what you get if you integrate Eqn. 1 between r1 and r2. If you insist in expressing this relationship in the form discussed in post #1, then you can write:
$$Q=-A_{LM}k\frac{(T_2-T_1)}{(r_2-r_1)}$$where $$A_{LM}=2\pi \frac{(r_2-r_1)}{\ln{(r_2/r_1)}}L$$. $A_{LM}$ is a weighted average of the cross sectional areas at r2 and r1 (called the logarithmic mean area), and $-k(T_2-T_1)/(r_2-r_1)$ is the average heat flux over the region between r2 and r1.

5. Apr 27, 2017

### whatdoido

Where does $Q= \frac {\Delta T}{R}$ come from then? The units don't add up without area $W= \frac{^0C}{^0Cm^2/W}$

6. Apr 27, 2017

### Staff: Mentor

The correct form of the equation for the heat flux q is $$q=-k\frac{dT}{dx}$$ where k is the thermal conductivity with units of $W/(m\ C)$, T is the temperature with units of C, and x is distance with units of m. So the units of the heat flux q are $W/m^2$. If we multiply this by the cross sectional area A (through which the heat is flowing), we obtain Q, which has units of W.