Question

Find the steady state temperature distribution in a circular annulus (shaded area) of inner radius 1 and outer radius 2 if the inner circle is held at $0^{\circ }$ and the outer circle has half its circumference at $0^{\circ }$ and half at ${100}^{\circ }.$ Hint: Don't forget the $r$ solutions corresponding to $k=0$.

Short answer

The temperature distribution in the annulus is a combination of simple radial functions adjusted by piecewise boundary conditions. Expect a smooth interpolation between boundary conditions along radial lines.

Step-by-step solution

Identify the Problem

We need to find the steady-state temperature distribution within a circular annulus with specified boundary conditions. The inner radius is 1, and the outer radius is 2. The inner boundary temperature is held at 0°C, and the outer boundary has half its circumference at 0°C and the other half at 100°C.

Set Up the Governing Equation

The steady-state temperature distribution in a cylindrical coordinate system is governed by Laplace’s equation for radial symmetry: $$\frac{{\partial }^{2}T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}=0$$

Solve the Differential Equation

The general solution to this differential equation is $$T(r)=A\ln r+B$$ where A and B are constants determined by boundary conditions.

Apply Inner Boundary Condition

The inner boundary condition is given as: $$T(1)=0°C.$$ Substitute r=1 in the general solution: $$0=A\ln (1)+B$$ Since $\ln (1)=0$, we have: $$B=0$$

Consider the Outer Boundary Condition

The outer boundary condition is more complex as half the circumference at radius 2 is at 0°C and the other half at 100°C. Due to the linearity of Laplace's equation, the solution to the temperature can be split into two problems: one where the boundary condition is 0°C everywhere and another where it is 0°C on half the boundary and 100°C on the other half.

Construct the Temperature Solution Using Superposition

The final temperature distribution needs to solve both parts of the boundary conditions. However, due to symmetry, we can approximate using a simple linear interpolation between 0°C and 100°C at the radius 2 boundary. This is generally valid due to equal partition boundary conditions.

Superimpose Solutions to Determine Constants

For simplicity and without loss of generality, this linear assumption gives us: For 0 < θ < π or half boundary at 0°C: T(2,θ) = 0°C For π < θ < 2π or other half boundary at 100°C: T(2,θ) = 100°C

Key concepts

Laplace's equation

In this problem, we need the steady state temperature distribution in the annular region. This requires solving Laplace's equation in the cylindrical coordinate system. Laplace's equation is a second-order partial differential equation (PDE) given by: $$abl{a}^{2}T=0$$.
For cylindrical coordinates, it translates to: $$\text{Missing argument for frac}$$
This means we only consider the radial part as it fits our problem setup, ignoring theta and z components.
The steady state implies there are no changes in temperature over time, making Laplace's equation relevant.

Boundary Conditions

Boundary conditions are crucial to solving PDEs like Laplace's equation. They tell us the values the solution must take at the boundaries of the domain. For our problem:

• Inner boundary: on the circle of radius 1, temperature is held at 0°C. So, T(r=1, θ) = 0.
• Outer boundary: radius 2 temperature varies. Half of its circumference is 0°C and the other half is 100°C. So, T(r=2, θ) = 0°C for 0 < θ < π and T(r=2, θ) = 100°C for π < θ < 2π.

Applying these will help to find the constants in our general solution.

Cylindrical Coordinate System

The cylindrical coordinate system $r,\theta ,\text{and}z$ simplifies problems with radial symmetry. For this problem, we take advantage of the radial nature.

• r: radial distance from the center.
• θ: the angular coordinate around the axis.
• z: the height along the cylinder's axis (though we're ignoring z as our problem is 2D).

Using it, our Laplace equation simplifies to functions of r and θ, ideal for representing a circular annulus.

Superposition Principle

The superposition principle states that if a problem can be broken into simpler problems, the solution is the sum of individual solutions. This principle applies to our Laplace problem due to linearity.
Two parts:

• One problem where all boundary is 0°C, giving T_1(r, θ=0).
• Another where boundary varies, half 0°C half 100°C, giving T_2(r,θ).

The combined solution T(r,θ) = T_1 + T_2.
This simplifies calculations and helps determine boundary behavior quickly.