Thick or infinitely thin. In these cases, we cannot use the lumped capacitance model or semi-infinite solid approximations and must instead use the heat diffusion equation to model the transient response. Let’s consider the case of a plane wall of thickness 2L which is suddenly subjected to convection at its surface. (a) Write the heat diffusion equation for the plane wall where only one spatial coordinate is needed to describe the internal temperature distribution. Make the assumptions that there is no internal energy generation and that the thermal conductivity is constant. (b) Initially, the plane wall is at a constant temperature, Ti. At t > 0, both surfaces of the plane wall are subjected to convection with a convection heat transfer coefficient, h, and temperature T… Write the initial condition and two boundary conditions. Each individual heat transfer problem will depend on different physical parameters such as the exact values of Ti, To, h, k, L, P, and Cp. In the next two steps, we will define some new quantities which serve to highlight the similarities that exist between different heat transfer problems which have the same geometry. (c) Write your differential equation (a) and initial/boundary conditions (b) in terms of the new variables, θ and θi where θ = T – T… and θi = Ti – T… (d) Use the definitions below and substitute into the differential equation, initial condition, and boundary conditions to further simplify the problem (this is like what you did in part (c) when you substituted in the temperature difference, θ for T). You should end up with everything only dependent on the dimensionless quantities θ, x, t, and Bi. θ = θ/θ x = x/L t* = αt/L^2 (note: t* in this definition is equal to the Fourier number) Bi = hL/k By nondimensionalizing the problem (the name for what you just did), we can see that for a prescribed geometry, the transient temperature distribution is a universal function of a dimensionless position x, a dimensionless time t (aka the Fourier number), and the Biot number Bi.
In many cases, a material cannot be approximated as either infinitely thick or infinitely thin. In these cases, we cannot use the lumped capacitance model or semi-infinite solid approximations and must instead use the heat diffusion equation to model the transient response. Let’s consider the case of a plane wall of thickness 2L which is suddenly subjected to convection at its surface. (a) Write the heat diffusion equation for the plane wall where only one spatial coordinate is needed to describe the internal temperature distribution. Make the assumptions that there is no internal energy generation and that the thermal conductivity is constant. (b) Initially, the plane wall is at a constant temperature, Ti. At t > 0, both surfaces of the plane wall are subjected to convection with a convection heat transfer coefficient, h, and temperature T… Write the initial condition and two boundary conditions. Each individual heat transfer problem will depend on different physical parameters such as the exact values of Ti, To, h, k, L, P, and Cp. In the next two steps, we will define some new quantities which serve to highlight the similarities that exist between different heat transfer problems which have the same geometry. (c) Write your differential equation (a) and initial/boundary conditions (b) in terms of the new variables, θ and θi where θ = T – T… and θi = Ti – T… (d) Use the definitions below and substitute into the differential equation, initial condition, and boundary conditions to further simplify the problem (this is like what you did in part (c) when you substituted in the temperature difference, θ for T). You should end up with everything only dependent on the dimensionless quantities θ, x, t, and Bi. θ = θ/θ x = x/L t* = αt/L^2 (note: t* in this definition is equal to the Fourier number) Bi = hL/k By nondimensionalizing the problem (the name for what you just did), we can see that for a prescribed geometry, the transient temperature distribution is a universal function of a dimensionless position x, a dimensionless time t (aka the Fourier number), and the Biot number Bi.
The Correct Answer and Explanation is:
(a) Heat Diffusion Equation for a Plane Wall
For a plane wall with one-dimensional heat conduction and no internal heat generation, the transient heat diffusion equation is:∂T∂t=α∂2T∂x2\frac{\partial T}{\partial t} = \alpha \frac{\partial^2 T}{\partial x^2}∂t∂T=α∂x2∂2T
Here:
- TTT is the temperature as a function of position xxx and time ttt
- α=kρcp\alpha = \frac{k}{\rho c_p}α=ρcpk is the thermal diffusivity
- kkk is thermal conductivity
- ρ\rhoρ is density
- cpc_pcp is specific heat
- x∈[−L,L]x \in [-L, L]x∈[−L,L]
(b) Initial and Boundary Conditions
- Initial Condition (at t=0t = 0t=0):
T(x,0)=Tifor −L≤x≤LT(x, 0) = T_i \quad \text{for } -L \le x \le LT(x,0)=Tifor −L≤x≤L
- Boundary Conditions (for t>0t > 0t>0):
At x=±Lx = \pm Lx=±L, convection occurs with ambient temperature T∞T_\inftyT∞ and heat transfer coefficient hhh:−k∂T∂x∣x=±L=h(T(x=±L,t)−T∞)-k \left. \frac{\partial T}{\partial x} \right|_{x = \pm L} = h (T(x = \pm L, t) – T_\infty)−k∂x∂Tx=±L=h(T(x=±L,t)−T∞)
(c) Define θ and Rewrite Equations
Let:θ(x,t)=T(x,t)−T∞,θi=Ti−T∞\theta(x, t) = T(x, t) – T_\infty, \quad \theta_i = T_i – T_\inftyθ(x,t)=T(x,t)−T∞,θi=Ti−T∞
Then the heat equation becomes:∂θ∂t=α∂2θ∂x2\frac{\partial \theta}{\partial t} = \alpha \frac{\partial^2 \theta}{\partial x^2}∂t∂θ=α∂x2∂2θ
With:
- Initial condition: θ(x,0)=θi\theta(x, 0) = \theta_iθ(x,0)=θi
- Boundary condition: −k∂θ∂x∣x=±L=hθ(x=±L,t)-k \left. \frac{\partial \theta}{\partial x} \right|_{x = \pm L} = h \theta(x = \pm L, t)−k∂x∂θx=±L=hθ(x=±L,t)
(d) Nondimensionalization
Define:θ∗=θθi,x∗=xL,t∗=αtL2,Bi=hLk\theta^* = \frac{\theta}{\theta_i}, \quad x^* = \frac{x}{L}, \quad t^* = \frac{\alpha t}{L^2}, \quad Bi = \frac{hL}{k}θ∗=θiθ,x∗=Lx,t∗=L2αt,Bi=khL
Substitute into the diffusion equation:∂θ∗∂t∗=∂2θ∗∂x∗2\frac{\partial \theta^*}{\partial t^*} = \frac{\partial^2 \theta^*}{\partial x^{*2}}∂t∗∂θ∗=∂x∗2∂2θ∗
Initial condition becomes:θ∗(x∗,0)=1\theta^*(x^*, 0) = 1θ∗(x∗,0)=1
Boundary conditions:∂θ∗∂x∗∣x∗=±1=−Bi θ∗(x∗=±1,t∗)\left. \frac{\partial \theta^*}{\partial x^*} \right|_{x^* = \pm 1} = -Bi \, \theta^*(x^* = \pm 1, t^*)∂x∗∂θ∗x∗=±1=−Biθ∗(x∗=±1,t∗)
Explanation
A transient heat conduction problem in a plane wall of thickness 2L2L2L requires the one-dimensional heat diffusion equation when internal energy generation is absent and thermal conductivity is constant. The temperature profile is assumed to vary only in the xxx-direction, from the center to the surface. This allows formulation of the transient equation using partial derivatives in time and space.
Initially, the entire wall is at a uniform temperature TiT_iTi. Once the surface is exposed to convection, heat exchange begins with the surroundings at temperature T∞T_\inftyT∞ through a coefficient hhh. This introduces boundary conditions relating the conductive heat flux to convective heat loss or gain at the wall surfaces.
To make the problem dimensionless and more universally applicable, a temperature difference variable θ=T−T∞\theta = T – T_\inftyθ=T−T∞ is introduced. This simplifies the boundary and initial conditions, eliminating explicit reference to ambient temperature and focusing on relative changes.
The next step introduces dimensionless groups: the dimensionless temperature θ∗\theta^*θ∗, spatial coordinate x∗=x/Lx^* = x/Lx∗=x/L, and time coordinate t∗=αt/L2t^* = \alpha t / L^2t∗=αt/L2, also known as the Fourier number. The Biot number Bi=hL/kBi = hL/kBi=hL/k characterizes the ratio of convective to conductive resistance and is essential in evaluating the importance of surface heat transfer compared to internal conduction.
Recasting the equations in this form shows that for any wall with the same geometry, the temperature evolution is governed solely by these dimensionless parameters. This nondimensional approach enables universal solutions that can be applied to various physical situations by scaling the dimensionless results.
