圆柱坐标系下圆柱热方程求解：稳态温度分布与传热量计算

Alright, let's work through this steady-state heat conduction problem in cylindrical coordinates step by step—this is a classic separation of variables application, so we'll break it down clearly.

Problem Setup #

First, let's list out all given parameters and constraints to keep things organized:

• Geometry: Inner radius a = 2.5 cm = 0.025 m, outer radius b = 3 cm = 0.03 m
• Material Property: Copper thermal conductivity K = 400 W/(m·K)
• Boundary Conditions:
  • Inner surface (r = a): Constant temperature T(a, θ) = 20 ℃
  • Outer surface (r = b): Temperature varies with angle: T(b, θ) = 50 + 10cosθ ℃
• Assumptions: Steady-state, no internal heat generation, and temperature is uniform along the pipe's length (z-direction), so we only need to solve for T(r, θ) in 2D cylindrical coordinates.

The governing equation is the Laplace equation for heat conduction:

∇²T = (1/r)∂/∂r(r∂T/∂r) + (1/r²)∂²T/∂θ² = 0

Separation of Variables Solution #

We assume a separable solution of the form T(r, θ) = R(r)Θ(θ). Plugging this into the Laplace equation splits it into two ordinary differential equations:

1. Angular ODE (θ-direction) #

Since temperature must be periodic with period 2π, we get:

• For n = 0: Θ₀(θ) = A₀ (constant)
• For n ≥ 1: Θₙ(θ) = Aₙcosnθ + Bₙsinnθ

2. Radial ODE (r-direction) #

This is an Euler equation, with solutions:

• For n = 0: R₀(r) = C₀ + D₀lnr
• For n ≥ 1: Rₙ(r) = Cₙrⁿ + Dₙr⁻ⁿ

3. Superpose Solutions & Apply Boundary Conditions #

Combining the solutions, we get:

T(r, θ) = C₀ + D₀lnr + Σₙ=1^∞ (Cₙrⁿ + Dₙr⁻ⁿ)(Aₙcosnθ + Bₙsinnθ)

Looking at the outer surface boundary condition, there are no sinnθ terms, so all Bₙ = 0. Now we match coefficients using both boundary conditions:

Constant Term (n=0)

From T(a, θ) = 20 and T(b, θ) = 50:

20 = C₀ + D₀lna
50 = C₀ + D₀lnb

Solving these gives:

D₀ = 30 / ln(b/a)
C₀ = 20 - 30lna/ln(b/a)

n=1 Term

From T(a, θ) = 20 (no cosθ term):

0 = C₁a + D₁/a → D₁ = -C₁a²

From T(b, θ) = 50 + 10cosθ:

10 = C₁b + D₁/b

Substitute D₁ and solve:

C₁ = 10b/(b² - a²)
D₁ = -10a²b/(b² - a²)

n ≥ 2 Terms

All coefficients Cₙ = Dₙ = 0 (since there are no higher-order cosine terms in the boundary conditions).

Final Temperature Distribution #

Putting it all together, the steady-state temperature is:

T(r, θ) = 20 + 30 * ln(r/a)/ln(b/a) + [10b(r - a²/r)/(b² - a²)]cosθ

Calculate Heat Power per Meter of Pipe #

The heat power Q per meter of pipe length is found by integrating the radial heat flux over the inner surface. Using Fourier's law, q_r = -K∂T/∂r, so:

Q = ∫₀^2π q_r(r=a) * a dθ = -K a ∫₀^2π (∂T/∂r)|_{r=a} dθ

First compute ∂T/∂r:

∂T/∂r = 30/(r ln(b/a)) + 10b/(b² - a²)(1 + a²/r²)cosθ

At r = a:

(∂T/∂r)|_{r=a} = 30/(a ln(b/a)) + 20b/(b² - a²)cosθ

When we integrate over 0 to 2π, the cosine term vanishes (since ∫₀^2π cosθ dθ = 0). This leaves:

Q = -60πK / ln(b/a)

The negative sign indicates heat flows from the outer surface to the inner surface (into the water). The magnitude of the heat power is:

|Q| = 60πK / ln(b/a)

Plugging in the numerical values (ln(0.03/0.025) ≈ 0.1823):

|Q| ≈ 60 * π * 400 / 0.1823 ≈ 413,600 W/m ≈ 413.6 kW/m

Key Notes #

• The temperature distribution has two parts: an axisymmetric radial component (from the constant temperature difference) and a periodic angular component (from the outer surface's cosine temperature variation).
• The periodic angular part doesn't contribute to the total heat power because it averages out over the full circumference—only the steady radial temperature gradient drives net heat flow into the water.

内容的提问来源于stack exchange，提问作者Future Math person