均匀生热环形固体稳态下内外壁温是否必然相等的求证问询

Great question—let’s break this down step by step, starting from the core equations governing steady-state heat transfer in your ring-shaped (cylindrical) solid.

First, the one-dimensional steady-state heat equation for radial conduction with uniform internal heat generation $\dot{q}$ is:
$$\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) + \frac{\dot{q}}{k} = 0$$
where $k$ is the solid’s thermal conductivity, $r$ is radial position, and $T$ is temperature.

Integrating the Equation to Find Temperature Distribution #

We can solve this equation by integrating twice:

1. First integration:
   $$r\frac{dT}{dr} = -\frac{\dot{q}}{2k}r^2 + C_1$$
   Rearranged to get the temperature gradient:
   $$\frac{dT}{dr} = -\frac{\dot{q}}{2k}r + \frac{C_1}{r}$$
2. Second integration gives the temperature profile:
   $$T(r) = -\frac{\dot{q}}{4k}r^2 + C_1\ln r + C_2$$
   Here, $C_1$ and $C_2$ are constants determined by your boundary conditions.

The Key: Boundary Conditions Don’t Require Equal Wall Temperatures #

Your TA focused on the case where $T_i = T_o$ (inner and outer wall temperatures equal) because it simplifies the math—but this is a special case, not a requirement for steady state.

Steady state only demands that the total heat generated inside the solid equals the total heat removed by the two cooling streams. Mathematically, that’s:
$$\dot{q} \cdot \pi(r_o^2 - r_i^2)L = q_i'' \cdot 2\pi r_i L + q_o'' \cdot 2\pi r_o L$$
where $r_i/r_o$ are inner/outer radii, $L$ is the ring’s length, and $q_i''/q_o''$ are the heat fluxes at the inner/outer walls.

If you set $T_i \neq T_o$, you can still solve for a valid steady-state temperature distribution:

• Plug $T(r_i) = T_i$ and $T(r_o) = T_o$ into the temperature profile equation to get a system of two equations.
• Solve for $C_1$ and $C_2$:
  $$C_1 = \frac{(T_o - T_i) + \frac{\dot{q}}{4k}(r_o^2 - r_i^2)}{\ln(r_o/r_i)}$$
  $$C_2 = T_i + \frac{\dot{q}}{4k}r_i^2 - C_1\ln r_i$$
  This gives a fully valid, non-symmetric temperature profile that satisfies both the heat equation and energy conservation.

Why the $T_i = T_o$ Case Is Taught #

That scenario is just easier to demonstrate: it leads to a clean, symmetric solution where the temperature gradient balances out the heat generation in a way that’s intuitive for students. But it’s by no means the only possible steady-state condition.

In short: Steady state does not require inner and outer wall temperatures to be equal. The only hard requirement is that the total heat removed matches the total heat generated. You can have different wall temperatures, different heat fluxes, or a mix—all as long as energy is conserved.

内容的提问来源于stack exchange，提问作者suguri