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In a hollow cylinder made of some uniform material with thermal conductivity k, length L and inner radius a and outer radius b, the total heat current (dQ/dt) has a constant value in the direction of heat transfer across the wall of the cylinder. How is this possible?

Supposing the temperature at a is greater than that at b, the radius will increase in the direction of heat transfer. This will increase the area through which heat is transferred, so why doesn't dQ/dt also increase?

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    $\begingroup$ What do mean by "throughout the cylinder"? $\endgroup$
    – Bob D
    Commented Apr 23, 2022 at 20:19
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    $\begingroup$ @BobD in the direction of heat transfer $\endgroup$
    – Piksiki
    Commented Apr 23, 2022 at 20:20
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    $\begingroup$ So do you mean heat transfer across the wall of the cylinder? $\endgroup$
    – Bob D
    Commented Apr 23, 2022 at 20:23
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    $\begingroup$ @BobD Yes that's what I mean $\endgroup$
    – Piksiki
    Commented Apr 23, 2022 at 20:23

2 Answers 2

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This will increase the area through which heat is transferred, so why doesn't dQ/dt also increase?

Because the heat flux, $\dot Q/A$ (watts/m$^2$), decreases with increasing radius.

See FIG 1 below. Assuming the inner wall is maintained at a constant temperature $T_a$ and the outer wall maintained at a constant temperature $T_b$, where $T_{a}\gt T_b$, then the direction of steady heat flow will be radial and perpendicular to the wall surfaces. The heat flow rate $\dot Q$ is given by the equation

$$\dot Q=\frac{2\pi kL(T_{a}-T_{b})}{\ln\biggl(\frac{r_b}{r_a}\biggr)}$$

Note that the thicker the cylinder wall (i.e., the greater the ratio of the outer to inner radius) not only will the magnitude of $\dot Q$ be lower, but the heat flux $\dot Q/A$ will also decrease. See FIG 2 which shows how the heat flux is uniform for a flat wall no matter how thick, but decreases for the cylindrical wall with increasing thickness.

Hope this helps.

enter image description here

enter image description here

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I've drawn a cross section of the hollow cylinder. Let's assume the heat is flowing outwards, though the argument applies whether the heat flow is outwards or inwards.

Cylinder

Consider the two surfaces at $r_1$ and $r_2$, and well call the total heat flow through those surfaces $q_1$ and $q_2$. If $q_1 > q_2$ then more heat is flowing into the region between the surfaces (the coloured region) from $r_1$ than is flowing out of it from $r_2$. That means the region must be getting hotter i.e. its temperature increases with time.

Conversely if $q_1 < q_2$ then less heat is flowing into the region between the surfaces (the coloured region) from $r_1$ than is flowing out of it from $r_2$. That means the region must be getting colder i.e. its temperature decreases with time.

But at equilibrium the temperatures are constant and don't change with time. Hence the only way the shell can be at equilibrium is if $q_1 = q_2$.

The heat flows can be different through $r_1$ and $r_2$, but only if the system is not at equilibrium i.e. if its temperatures are changing with time.

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