Problem 425 - Calculus Volume 3

Question

For the following exercises, Fourier's law of heat transfer states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region \(D\) is given. Use the divergence theorem to find net outward heat \(\quad\) flux \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S=-k \iint_{S} \nabla T \cdot \mathbf{N} d S\) across the boundary \(S\) of \(D,\) where \(k=1\). \(T(x, y, z)=100+e^{-z}\) \(D=\mid(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\}\)

Step-by-Step Solution

Answer

The net outward heat flux across the boundary is \( 1 - \frac{1}{e} \).

Step 1: Understanding the Mathematical Principles

The heat flow vector \( \mathbf{F} \) is given by Fourier's law, \( \mathbf{F} = -k abla T \). The divergence theorem relates surface integrals to volume integrals, allowing us to convert the surface integral of the flux on the boundary \( S \) into a volume integral of the divergence within \( D \). The net heat flux is \( \iint_S \mathbf{F} \cdot \mathbf{N} \, dS = \iiint_D abla \cdot \mathbf{F} \, dV \).

Step 2: Calculate the Gradient of the Temperature Function

To find the heat flow vector, we must first calculate the gradient of the temperature function, \( T(x, y, z) = 100 + e^{-z} \). The gradient \( abla T \) is given by the partial derivatives with respect to \( x, y, \) and \( z \). Since \( T \) does not depend on \( x \) or \( y \), the partial derivatives \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \) are both zero. The partial derivative with respect to \( z \) is \( \frac{\partial T}{\partial z} = -e^{-z} \). Thus, \( abla T = \begin{bmatrix} 0 \ 0 \ -e^{-z} \end{bmatrix} \).

Step 3: Substitute Gradient into Fourier's Law

Using \( k = 1 \), substitute \( abla T \) into Fourier's law: \( \mathbf{F} = -abla T = \begin{bmatrix} 0 \ 0 \ e^{-z} \end{bmatrix} \).

Step 4: Calculate the Divergence of \( \mathbf{F} \)

The divergence of \( \mathbf{F} \) is calculated as the dot product of \( abla \) with \( \mathbf{F} \). Since \( \mathbf{F} = \begin{bmatrix} 0 \ 0 \ e^{-z} \end{bmatrix} \), we have \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x} (0) + \frac{\partial}{\partial y} (0) + \frac{\partial}{\partial z} (e^{-z}) = -e^{-z} \).

Step 5: Apply the Divergence Theorem to Convert Surface Integral to Volume Integral

According to the divergence theorem, \( \iint_S \mathbf{F} \cdot \mathbf{N} \, dS = \iiint_D (abla \cdot \mathbf{F}) \, dV \). Substitute the divergence \( -e^{-z} \) and evaluate over the region \( D \), \( 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1 \).

Step 6: Compute the Volume Integral

The volume integral \( \iiint_D -e^{-z} \, dV \) separates into: \( \int_0^1 \int_0^1 \int_0^1 -e^{-z} \, dz \, dy \, dx \). First integrate with respect to \( z \): \( \int_0^1 -e^{-z} \, dz = \left.[-e^{-z}]_0^1 = (-e^{-1}) - (-1) = 1 - \frac{1}{e} \). The integrals with respect to \( y \) and \( x \) are simple over the interval \([0, 1]\), each contributing a factor of 1, since \( \int_0^1 1 \, dy = 1 \) and \( \int_0^1 1 \, dx = 1 \).

Step 7: Calculate the Net Outward Heat Flux

Putting it all together, the complete integral is \( 1 (\text{from } x) \times 1 (\text{from } y) \times (1 - \frac{1}{e}) \). Thus, the net outward heat flux across boundary \( S \) is \( 1 - \frac{1}{e} \).

Key Concepts

Divergence Theorem

The Divergence Theorem is a powerful tool in vector calculus. It helps us convert surface integrals into volume integrals. This theorem is particularly useful in fields like fluid dynamics and electromagnetism. In the context of heat transfer, it simplifies the calculation of net heat flow across a closed surface. The theorem states that for a vector field \( \mathbf{F} \), the surface integral over a closed surface \( S \) can be related to a volume integral over the region \( D \) enclosed by \( S \). The relationship is given by:

\[ \iint_S \mathbf{F} \cdot \mathbf{N} \, dS = \iiint_D abla \cdot \mathbf{F} \, dV\]
In simple terms, this means that the total "outflow" of a vector field through a closed surface is equal to the "sum of sources" within the volume. In our problem, this allows us to find the heat flux through the boundary from the divergence within the region.

Heat Flux

Heat flux refers to the rate of heat energy transfer per unit area. This transfer occurs across a surface. In Fourier's Law, the heat flux vector \( \mathbf{F} \) describes the direction and magnitude of heat transfer. The law tells us that heat flows from warmer to cooler regions. The formula for the heat flow vector is:

\[ \mathbf{F} = -k abla T \]
Where:

• \( \mathbf{F} \) is the heat flux vector
• \( k \) is the thermal conductivity constant of the material
• \( abla T \) is the gradient of the temperature function

A negative sign indicates that heat moves opposite to the direction of the temperature gradient. This principle underlies the exercise, where ultimately the net heat flux is found through integration.

Gradient of Temperature Function

The gradient of a temperature function, denoted as \( abla T \), plays a crucial role in describing heat flow. It is a vector that points in the direction of the greatest rate of increase of the temperature function. Its magnitude represents the rate of change in this direction.

For the given temperature function \( T(x, y, z) = 100 + e^{-z} \), the gradient is calculated using partial derivatives. Since \( T \) doesn't depend on \( x \) or \( y \), the partial derivatives \( \frac{\partial T}{\partial x} \) and \( \frac{\partial T}{\partial y} \) are zero, yielding:
\[ abla T = \begin{bmatrix} 0 \ 0 \ -e^{-z} \end{bmatrix}\]This result informs us that the temperature decreases most rapidly in the negative \( z \)-direction. Substituting the gradient into Fourier's Law gives us the heat flux vector, manifesting the direction and intensity of the heat transfer throughout the region. This understanding is vital for analyzing the given problem scenario.