Problem 32
Question
The heat diffusion equation Let \(T(t, x, y, z)\) be a function with continuous second derivatives giving the temperature at time \(t\) at the point \((x, y, z)\) of a solid occupying a region \(D\) in space. If the solid's heat capacity and mass density are denoted by the constants \(c\) and \(\rho,\) respectively, the quantity \(c \rho T\) is called the solid's heat energy per unit volume. a. Explain why - \(\nabla T\) points in the direction of heat flow. b. Let \(-k \nabla T\) denote the energy flux vector. (Here the constant \(k\) is called the conductivity.) Assuming the Law of Conservation of Mass with \(-k \nabla T=\mathbf{v}\) and \(c \rho T=p\) in Exercise \(31,\) derive the diffusion (heat) equation $$\frac{\partial T}{\partial t}=K \nabla^{2} T$$ where \(K=k /(c \rho)>0\) is the diffusivity constant. (Notice that if \(T(t, x)\) represents the temperature at time \(t\) at position \(x\) in a uniform conducting rod with perfectly insulated sides, then \(\nabla^{2} T=\partial^{2} T / \partial x^{2}\) and the diffusion equation reduces to the one-dimensional heat equation in Chapter 14 's Additional Exercises.)
Step-by-Step Solution
VerifiedKey Concepts
Temperature Gradient
Energy Flux
Thermal Conductivity
- Higher values of k imply better heat transfer
- Metals are typically very good at conducting heat
- Insulators have low thermal conductivity, preventing energy flow