Question

Starting with an energy balance on a rectangular volume element, derive the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation.

Short answer

Question: Derive the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation. Answer: The one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation can be derived as: \(\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial x^2}\) This equation shows how the temperature distribution within the plane wall changes with time due to heat conduction.

Step-by-step solution

Define the volume element

Let's define a rectangular volume element in the plane wall with width dx, height h, and length L. The volume of this element is V = h * L * dx.

Apply energy conservation principles

Perform an energy balance on the volume element. The rate at which energy enters the element minus the rate at which energy exits the element equals the rate at which energy is stored within the element.

Write energy balance equation

Let q(x) be the heat flux entering the element at x, and q(x+dx) be the heat flux leaving the element at x+dx. There is no heat generation within the element. Then, the energy balance equation can be written as: \(-q(x)A + q(x+dx)A = \frac{\partial}{\partial t}(V \rho c_p T)\) where A is the cross-sectional area of the element (h * L), ρ is the density, c_p is the specific heat, and T is the temperature.

Apply Fourier's Law

Since we are dealing with conduction heat transfer, we can apply Fourier's Law, which states that the heat flux is proportional to the temperature gradient and the thermal conductivity. For the plane wall with constant thermal conductivity k, we have: \(q(x) = -k \frac{\partial T}{\partial x}\) and \(q(x + dx) = -k \frac{\partial T(x+dx)}{\partial x}\)

Substitute Fourier's Law expressions into the energy balance equation

Now, substitute the Fourier's Law expressions from Step 4 into the energy balance equation from Step 3: \[kA \frac{\partial T}{\partial x} - kA \frac{\partial T(x+dx)}{\partial x} = hL \rho c_p \frac{\partial T}{\partial t} dx\]

Expand the Taylor series and simplify

Next, we expand the second term on the left side of the equation using Taylor's expansion: \(kA \frac{\partial T}{\partial x} - kA \left( \frac{\partial T}{\partial x} + \frac{\partial^2 T}{\partial x^2} dx \right) = hL \rho c_p \frac{\partial T}{\partial t} dx\) Simplify the equation by canceling terms and dividing by \(hL dx\): \(- k \frac{\partial^2 T}{\partial x^2} = \rho c_p \frac{\partial T}{\partial t}\)

Rearrange to get the one-dimensional heat conduction equation

Finally, we rearrange the equation to get the one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and no heat generation: \(\frac{\partial T}{\partial t} = \frac{k}{\rho c_p} \frac{\partial^2 T}{\partial x^2}\) This is the desired equation, which shows how the temperature distribution within the plane wall changes with time due to heat conduction.