Question

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) \(T(x, y, z)=100 e^{-x-y} ; S\) consists of the faces of the cube \(|x| \leq 1\) \(|y| \leq 1,|z| \leq 1\)

Short answer

Q: Calculate the outward flux of the heat flow vector field across the faces of a cube with side length 2 centered at the origin based on a temperature distribution \(T(x, y, z) = 100e^{-x-y}\) and a thermal conductivity \(k=1\). A: The outward flux across the faces of the cube is approximately -218.39.

Step-by-step solution

Finding the gradient of the temperature function

First, we need to find the gradient of the given temperature function, \(T(x, y, z) = 100e^{-x-y}\). Recall that the gradient operator, denoted by \(\nabla\), is defined as follows: \(\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)\). To find the gradient of \(T\), we'll take the partial derivatives of the function with respect to \(x\), \(y\), and \(z\). The result is: \(\nabla T(x, y, z) = \left(-100e^{-x-y}, -100e^{-x-y}, 0\right)\)

Computing the heat flow vector field

Next, we'll compute the heat flow vector field, \(\mathbf{F}\), by taking the negative of the gradient and multiplying by the constant \(k\). Since we're told to assume \(k=1\), the heat flow vector field is simply \(\mathbf{F}=-\nabla T\). Using the gradient we found in Step 1, we get: \(\mathbf{F}(x, y, z) = \left(100e^{-x-y}, 100e^{-x-y}, 0\right)\)

Evaluating the outward flux

Finally, we'll calculate the outward flux of the heat flow vector field across the surface S, which is given by the following integral: \(\oint_S \mathbf{F} \cdot \mathbf{n} \, dS\) Here, \(\mathbf{n}\) is the unit outward normal vector of the surface S. Since S is the faces of a cube, the outward normal vectors for the six faces are: \((\pm 1, 0, 0)\), \((0, \pm 1, 0)\), and \((0, 0, \pm 1)\). Now, we can evaluate the integral for each face, and then sum the results. For example, on the face \(x=-1\), we have: \(\oint_{S_{x=-1}} \mathbf{F} \cdot (-1, 0, 0) \, dS = \int_{-1}^{1} \int_{-1}^{1} 100e^{-(-1)-y}(-1) \, dy\,dz\) Similar integrals can be computed for the other faces of the cube. Note that since the temperature function does not depend on \(z\), the contributions from the faces \(z = \pm 1\) will be zero. After calculating all the integrals and summing the results, we get: Outward flux \(= \int_{-1}^{1} \int_{-1}^{1} [100e^{-(-1)-y}(-1) + 100e^{-1-y}(1) - 100e^{-x-(-1)}(+1) + 100e^{-x-1}(-1)] \, dy\,dx\) After evaluating the integral, the final outward flux is approximately \(-218.39\).

Key concepts

Gradient Operator

The gradient operator, often denoted by \( abla \), plays a crucial role in the field of vector calculus. It is essentially a vector that consists of the first-order partial derivatives of a scalar function. In mathematical terms, for a function \( T(x, y, z) \), the gradient operator is expressed as:

• \( abla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \).

This operator provides a way to compute the rate at which the function changes as you move in any direction in space.
The gradient vector points in the direction of the steepest ascent of the function, and its magnitude is the rate of increase in that direction.
In the given exercise, the gradient was computed for the temperature function \( T(x, y, z) = 100e^{-x-y} \). By applying the gradient operator, we found:

• \( abla T(x, y, z) = \left(-100e^{-x-y}, -100e^{-x-y}, 0\right) \).

This shows that the temperature changes most rapidly in the directions of \( x \) and \( y \), while there is no change along the \( z \)-axis.

Heat Flow Vector Field

The heat flow vector field, represented as \( \mathbf{F} \), describes the transfer of heat in a medium. It is derived from the gradient of the temperature distribution. For conducting objects, the heat flow vector field is defined by:

• \( \mathbf{F} = -kabla T \),

where \( k \) is a positive constant characteristic of the material. It signifies the conductivity of the substance, although in our exercise we consider \( k=1 \).
The negative sign indicates that heat flows from regions of higher temperature to regions of lower temperature—opposing the direction of the gradient.
Thus, for our temperature function and given \( k=1 \), the heat flow vector field is:

• \( \mathbf{F} = (100e^{-x-y}, 100e^{-x-y}, 0) \).

This vector field shows that heat flows along the negative \( x \) and \( y \) directions, highlighting where and how the temperature decreases most.

Outward Flux

The concept of outward flux is pivotal when considering the transfer of heat across a surface. Outward flux measures the total amount of the vector field that passes through a given surface in an outward direction.
The mathematical tool to compute the outward flux across a surface \( S \) is the surface integral:

• \( \oint_S \mathbf{F} \cdot \mathbf{n} \, dS \),

where \( \mathbf{n} \) is the unit outward normal vector to the surface.
In the exercise, the surface \( S \) is a cube, and each face has its own outward normal vector. The flux calculation involves integrating over each face and adding these contributions together.
For the cube given, the faces are aligned with coordinate axes, simplifying the calculation of the outward flux to specific integrals. An essential insight here is recognizing that the flux contribution from faces perpendicular to one direction is null when the temperature function does not involve that variable.

Temperature Distribution

Temperature distribution, represented as \( T(x, y, z) \), indicates how temperature varies throughout a given medium. In our exercise, this was given as:

• \( T(x, y, z) = 100e^{-x-y} \).

This distribution reveals that the temperature depends exponentially on \( x \) and \( y \).
Such a setup suggests that as either \( x \) or \( y \) increases, the temperature decreases, which is typical in cooling patterns.
The exponential term \( e^{-x-y} \) results in a rapid decrease of temperature as you move away from the origin in the \( x\)-\( y \)-plane.
Graphically, this temperature distribution might resemble a ‘hill,’ where the height diminishes swiftly as you move along \( x \) and \( y \) axes, offering insight into the thermal dynamics within the object.