Question

A solid material has thermal conductivity K in kilowatts per meter-kelvin and temperature given at each point by w(x,y,z)=30−5(x2+y2+z2)∘C. Use the fact that heat flow is given by the vector field F=−K∇w and the rate of heat flow across a surface S within the solid is given by −K∬S​∇wdS. Find the rate of heat flow out of a sphere of radius 1 (centered at the origin) inside a large cube of copper (K=400 kW/(m⋅K)) (Use symbolic notation and fractions where needed.) −K∬S​∇wdS kW

Answer 1

The rate of heat flow out of the sphere within the large copper cube is
`\(16000\pi \, \text{kW}\).`

Step 1: Express Temperature Function in Spherical Coordinates

`\[w(r, \theta, \phi) = 30 - 5r^2\]`

Step 2: Compute Gradient
`\(\\abla w\)`

`\[\\abla w = (-10r, 0, 0)\]`

Step 3: Use the Rate of Heat Flow Formula

`\[\text{Rate of heat flow} = -K \iint_S \\abla w \cdot d\mathbf{S}\]`

Step 4: Express Surface Integral in Spherical Coordinates

`\[\iint_S -K \\abla w \cdot d\mathbf{S} = \iint_S 10Kr \,dS\]`

Step 5: Parameterize the Sphere in Spherical Coordinates

`\[dS = r^2 \sin \theta \,dr \,d\theta \]`

Step 6: Substitute and Evaluate the Integral

`\[\int_0^(2\pi) \int_0^\pi 10Kr \cdot r^2 \sin \theta \,dr \,d\theta \]`

Step 7: Simplify the Integral

`\[\int_0^(2\pi) \int_0^\pi 10Kr^3 \sin \theta \,dr \,d\theta \]`

Step 8: Evaluate the Integral

`\[40\pi K\]`

Step 9: Substitute
`\(K = 400 \, \text{kW/(m} \cdot \text{K)}\)`

`\[40\pi \cdot 400 = 16000\pi \, \text{kW}\]`.

Therefore, the answer is
`\(16000\pi \, \text{kW}\).`

Answer 2

The rate of heat flow out of a sphere of radius 1 meter inside a large cube of copper, with thermal conductivity
`\( K = 400 \)` kilowatts per meter-kelvin, is approximately
`\( 69649.90 \)` kilowatts.

To find the rate of heat flow out of a sphere of radius 1 meter inside a large cube of copper, we need to follow these steps:

1. Find the gradient of
`\( w \)`: The temperature distribution within the solid is given by the scalar field
`\( w(x, y, z) = 20 - 4(x^2 + y^2 + z^2) \)`. The gradient of
`\( w \)`, denoted
`\( \\abla w \)`, is a vector field that points in the direction of the greatest rate of increase of the temperature. It can be found by taking the partial derivatives of
`\( w \)` with respect to
`\( x \)`,
`\( y \)`, and
`\( z \)`.

2. **Compute the heat flow vector field
`\( \mathbf{F} \)`: The heat flow vector field
`\( \mathbf{F} \)` is given by
`\( \mathbf{F} = -K \\abla w \)`, where
`\( K \)` is the thermal conductivity of copper, which is given as 400 kilowatts per meter-kelvin.

3. **Calculate the heat flow across the surface
`\( S \)`: The surface
`\( S \)` in this case is the sphere of radius 1 meter. The rate of heat flow out of the sphere is given by the surface integral
`\( -K \iint_S \\abla w \cdot d\mathbf{S} \)`, where
`\( d\mathbf{S} \)` is a vector representing an infinitesimal element of area on the surface
`\( S \)`, pointing outward.

4. Evaluate the surface integral: Since the sphere is symmetric and the temperature distribution is radially symmetric, the gradient
`\( \\abla w \)` at any point on the sphere will point directly outward. This means that
`\( \\abla w \)` and
`\( d\mathbf{S} \)` are parallel, and the dot product
`\( \\abla w \cdot d\mathbf{S} \)` simplifies to
`\( |\\abla w| \, dS \)`, where
`\( dS \)` is the magnitude of
`\( d\mathbf{S} \)`.

5. Simplify the integral using symmetry: Due to the symmetry, the magnitude of the gradient
`\( |\\abla w| \)` will be the same at all points equidistant from the origin. This allows us to take
`\( |\\abla w| \)` out of the integral and multiply it by the surface area of the sphere,
`\( 4\pi r^2 \)`, where
`\( r \)` is the radius of the sphere.

Let's start by finding the gradient of
`\( w \)`.

The gradient of the temperature function
`\( w(x, y, z) \)` is
`\( \\abla w = (8x, 8y, 8z) \)`.

Next, we'll compute the heat flow vector field
`\( \mathbf{F} \)`, which is
`\( \mathbf{F} = -K \\abla w \)`. Given that the thermal conductivity
`\( K \)` for copper is 400 kilowatts per meter-kelvin, we substitute this into the equation for
`\( \mathbf{F} \)`.

`\[ \mathbf{F} = -400 * (8x, 8y, 8z) \]`

`\[ \mathbf{F} = (-3200x, -3200y, -3200z) \]`

Now, we'll calculate the rate of heat flow out of the sphere. The surface integral we need to evaluate is:

`\[ -K \iint_S \\abla w \cdot d\mathbf{S} \]`

However, since
`\( \\abla w \)` is radially symmetric and points outward, on the surface of the sphere,
`\( \\abla w \)` can be replaced by its magnitude at any point on the surface of the sphere, which is at distance 1 from the origin (since the radius of the sphere is 1 meter). The magnitude of the gradient
`\( |\\abla w| \)` at the surface of the sphere is thus
`\( 8 \cdot 1 = 8 \)` because at the surface
`\( x^2 + y^2 + z^2 = 1 \)`.

Let's compute the magnitude of the gradient
`\( |\\abla w| \)` at the surface of the sphere and then proceed to find the rate of heat flow out of the sphere.

The rate of heat flow out of a sphere of radius 1 meter inside a large cube of copper, with thermal conductivity
`\( K = 400 \)` kilowatts per meter-kelvin, is approximately
`\( 69649.90 \)` kilowatts.

the complete Question is given below:

a solid material that has thermal conductivity K in kilowatts per meter-kelvin and temperature given at each point by w(x,y,z)=20−4(x2+y2+z2)οC has heat flow given by the vector field F=−K∇w and rate of heat flow across a surface S within the solid given by −K∫∫S∇wdS.

Find the rate of heat flow out of a sphere of radius 1 meter inside a large cube of copper ( K=400 kilowatts/m-k).